Solution:
The problem says "some", so not all cells must be occupied.
We start by doing casework on the column on the left. There can be 5,4,3,2, or 1 black chip. The same goes for white chips, so we will multiply by 2 at the end. There is $1$ way to select $5$ cells with black chips. Because of the 2nd condition, there can be no white, and the grid must be all black- $1$ way . There are $5$ ways to select 4 cells with black chips. We now consider the row that does not contain a black chip. The first cell must be blank, and the remaining $4$ cells have $2^4-1$ different ways($-1$ comes from all blank). This gives us $75$ ways. Notice that for 3,2 or 1 black chips on the left there is a pattern. Once the first blank row is chosen, the rest of the blank rows must be ordered similarly. For example, with 2 black chips on the left, there will be 3 blank rows. There are 15 ways for the first row to be chosen, and the following 2 rows must have the same order. Thus, The number of ways for 3,2,and 1 black chips is $10*15$, $10*15$, $5*15$. Adding these up, we have $1+75+150+150+75 = 451$. Multiplying this by 2, we get $\boxed{902}$.
~westwoodmonster
Note that the answer is equivalent to the number of ways to choose rows and columns that the white chips occupy, as once those are chosen, there is only one way to place the chips, and every way to place the chips corresponds to a set of rows and columns occupied by the white pieces.
If the white pieces occupy none of the rows, then because they don't appear on the board, they will not occupy any of the columns. Similar logic can be applied to show that if white pieces occupy all of the rows, they will also occupy all of the columns.
The number of sets of rows and columns that white can occupy are $2^{5} - 2 = 30$ each, accounting for the empty and full set.
So, including the board with 25 white pieces and the board with 25 black pieces, the answer is $30^{2}+2 = \boxed{902}$
Case 1: All chips on the grid have the same color.
In this case, all cells are occupied with chips with the same color.
Therefore, the number of configurations in this case is 2.
Case 2: Both black and white chips are on the grid.
Observation 1: Each colored chips must occupy at least one column and one row.
This is because, for each given color, there must be at least one chip. Therefore, all chips placed in the cells that are in the same row or the same column with this given chip must have the same color with this chip.
Observation 2: Each colored chips occupy at most 4 rows and 4 columns.
This directly follows from Observation 1.
Observation 3: For each color, if all chips in this color occupy columns with $x$-coordinates $\left\{ x_1, \cdots, x_m \right\}$ and rows with $y$-coordinates $\left\{ y_1, \cdots, y_n \right\}$, then every cell $\left( x, y \right)$ with $x \in \left\{ x_1, \cdots , x_m \right\}$ and $y \in \left\{ y_1, \cdots , y_n \right\}$ is occupied by a chip with the same color.
This is because, if there is any cell in this region occupied by a chip with a different color, it violates Condition 2.
If there is any cell in this region that is empty, then it violates Condition 3.
Observation 4: For each color, if all chips in this color occupy columns with $x$-coordinates $\left\{ x_1, \cdots, x_m \right\}$ and rows with $y$-coordinates $\left\{ y_1, \cdots, y_n \right\}$, then every cell $\left( x, y \right)$ with $x \notin \left\{ x_1, \cdots , x_m \right\}$ and $y \in \left\{ y_1, \cdots , y_n \right\}$, or $x \in \left\{ x_1, \cdots , x_m \right\}$ and $y \notin \left\{ y_1, \cdots , y_n \right\}$ is empty.
This is because, if there is any cell in this region occupied by a chip with a different color, it violates Condition 2.
Observation 5: For each color, if all chips in this color occupy columns with $x$-coordinates $\left\{ x_1, \cdots, x_m \right\}$ and rows with $y$-coordinates $\left\{ y_1, \cdots, y_n \right\}$, then every cell $\left( x, y \right)$ with $x \notin \left\{ x_1, \cdots , x_m \right\}$ and $y \notin \left\{ y_1, \cdots , y_n \right\}$ is occupied by chips with the different color.
This follows from Condition 3.
By using the above observations, the number of feasible configurations in this case is given by
\begin{align*}
\sum_{n=1}^4 \sum_{m=1}^4 \binom{5}{n} \binom{5}{m}
& = \left( \sum_{n=1}^4 \binom{5}{n} \right)
\left( \sum_{m=1}^4 \binom{5}{m} \right) \\
& = \left( 2^5 - 2 \right)^2 \\
& = 900 .
\end{align*}
Putting all cases together, the total number of feasible configurations is $2 + 900 =  \boxed{\textbf{(902) }}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Answer: 902

<｜begin▁of▁sentence｜><｜User｜>There is a collection of $25$ indistinguishable white chips and $25$ indistinguishable black chips. Find the number of ways to place some of these chips in the $25$ unit cells of a $5\times5$ grid such that:

each cell contains at most one chip
all chips in the same row and all chips in the same column have the same colour
any additional chip placed on the grid would violate one or more of the previous two conditions.<｜Assistant｜><think>

Okay, so I have this problem here about placing chips on a 5x5 grid. Let me try to parse it step by step. The problem states:

We have 25 white chips and 25 black chips, all indistinct except for their color. We need to place some of these chips into the 25 unit cells of a 5x5 grid following three conditions:

1. Each cell can contain at most one chip.
2. All chips in the same row and all chips in the same column must have the same color.
3. The placement must be maximal, meaning we can't add any more chips without violating the first two conditions.

And we need to find the number of such arrangements. Hmm, okay. Let me try to break this down.

First, since each row and each column must have chips of the same color, that suggests that each row is either all white, all black, or empty? Wait, but actually, the problem says "some of these chips," so rows and columns can be partially filled? Wait, no, hold on. Wait, the second condition says "all chips in the same row and all chips in the same column have the same colour." So, if a row has any chips, they must all be the same color. Similarly for columns. So each row can be either all white, all black, or empty. Similarly, each column can be all white, all black, or empty. But since the grid is 5x5, the rows and columns are interdependent.

But also, the third condition says that the placement is maximal, so you can't add any more chips without violating the first two conditions. That means that every empty cell must be in a row or column that is already filled with the opposite color? Wait, maybe not exactly, but that adding another chip would require that its row or column already has a different color. Hmm, perhaps.

Alternatively, the maximal condition means that for every empty cell, if you were to place a chip there, it would conflict with either its row or its column's existing color. So, for every empty cell (i,j), either row i is already filled with some color (so adding another chip would require it to be that color, but maybe the column j has a different color?), or column j is already filled with some color, so you can't add a chip there. Wait, actually, the problem says "any additional chip placed on the grid would violate one or more of the previous two conditions." The two conditions are: each cell has at most one chip (so you can't place a chip where there's already one), and all chips in the same row and column have the same color. So, if you try to place a chip in an empty cell, you have to check that the row and column of that cell don't already have chips of a different color. So, the maximal condition requires that every empty cell is in a row that already has chips of some color, or in a column that already has chips of some color, such that you can't place a chip of either color there without conflicting.

Alternatively, maybe the row or column is already filled with the opposite color? Hmm, perhaps.

Alternatively, maybe the maximal condition is equivalent to saying that every row and every column is either completely filled or completely empty? Wait, no, because if a row is completely filled with white chips, then you can't add any more, but if it's partially filled, then maybe you could add more? Wait, but the second condition requires that all chips in the same row must be the same color, so if a row has some white chips, you can't add a black chip, but you could add more white chips? Wait, but the problem says "each cell contains at most one chip," so once a cell is filled, you can't add another. Wait, but if a row has some white chips, you can still add more white chips in the remaining cells of the row, as long as they are empty. But the maximal condition would require that you can't add any more chips, so that would mean that every row that is not completely filled must have some constraint from the columns?

Hmm, this is getting a bit confusing. Let me try to rephrase the problem.

We need to place some chips (either white or black) in the grid such that:

1. No two chips in the same cell.

2. In every row, all chips are the same color (if any).

3. In every column, all chips are the same color (if any).

4. The arrangement is maximal, so you can't add any more chips without violating 1, 2, or 3.

So, to satisfy condition 4, for every empty cell (i,j), either:

- The row i already has chips of some color, so adding a chip of the opposite color would violate the row's color consistency, and adding a chip of the same color would require that the column j doesn't have a different color. Wait, actually, if the row i has, say, white chips, then to add a white chip in (i,j), the column j must not have any black chips. But if column j already has black chips, then you can't add a white chip there. Alternatively, if column j is empty or has white chips, then you could add a white chip. But since the arrangement is maximal, that must not be possible. Therefore, for every empty cell (i,j), either:

- Row i is already filled with some color (so you can't add another chip of that color, but could you add the opposite color? Wait, no, because if the row is filled with white, you can't add a black chip to the row, since the row must remain all white. Wait, actually, if the row already has some white chips, then adding a black chip would violate the row's color consistency. So, in order to prevent adding any chip to (i,j), either:

- The row i has chips of some color, so you can't add a different color, and the column j also has chips of some color, so you can't add a different color. Therefore, the cell (i,j) is blocked by both row and column? Or maybe just one of them?

Wait, actually, to prevent adding any chip (either color) to (i,j), the following must hold: if you try to place a white chip there, then either the row already has black chips, or the column already has black chips. Similarly, if you try to place a black chip there, then either the row already has white chips, or the column already has white chips. Therefore, to block both possibilities, the cell (i,j) must be in a row with some color and a column with the opposite color. For example, if row i is all white and column j is all black, then you can't place a white chip because column j is black, and you can't place a black chip because row i is white. Therefore, such a cell is blocked. Alternatively, if the row is white and the column is white, then you could place a white chip there, so that cell is not blocked. Wait, but if the row is already filled with white chips, then the cell is already occupied? Wait, no, the row could have some white chips but not all. Wait, but if the row has some white chips, then you can add more white chips as long as the column allows it. Hmm, this is getting a bit tangled.

Alternatively, maybe the maximal condition requires that every row and every column is either entirely filled or entirely empty? Because if a row is partially filled, then you could potentially add more chips of the same color, unless the columns interfere. Wait, but if a row is partially filled with white chips, then the columns of those chips must also be all white. Wait, no, the columns can have their own colors. Wait, actually, each column must also have all chips of the same color. So, for example, if in column j, there is a white chip in some row, then all chips in column j must be white. Therefore, if a column has any white chips, it can't have any black chips, and vice versa.

Therefore, the colors of the rows and columns are interdependent. Let me think of this as a matrix where each row is either white, black, or empty, and each column is also white, black, or empty, with the constraint that if a row is colored, then all its non-empty cells must be that color, and similarly for columns. Moreover, the color of a row and the color of a column must agree on their intersection if both are colored. Wait, actually, if a row is white and a column is black, their intersection cell cannot have a chip, because that cell would have to be both white and black. Therefore, in such a case, the intersection must be empty. So, the rows and columns form a kind of partition where their colors must agree where they intersect, or else the intersection is empty.

Hmm, perhaps the problem can be modeled by choosing a subset of rows and columns, assigning each a color, such that if a row and column are both assigned colors, they must be the same color. Wait, not exactly. Let me think again.

Suppose we have some rows colored (either white or black) and some columns colored (either white or black). The intersection of a colored row and a colored column must be of the same color. Additionally, any cell that is in a colored row or a colored column must be filled with the respective color, but if both the row and column are colored, they must agree. However, the problem allows some cells to be empty, but with the maximal condition. Wait, perhaps the maximal condition requires that every uncolored row and column can't be colored without conflict? Hmm, maybe another approach.

Alternatively, since the problem requires that the placement is maximal, that means that for every empty cell, you cannot place a chip of either color there without violating the row or column color constraints. Therefore, for each empty cell (i,j):

- If you try to place a white chip there, then either row i already has a black chip, or column j already has a black chip.

- Similarly, if you try to place a black chip there, then either row i already has a white chip, or column j already has a white chip.

Therefore, for each empty cell (i,j), the row i and column j must be "blocked" in both colors. That is, either:

- Row i is black and column j is white, so you can't place white (because column j is white? Wait, no. Wait, if column j is white, then placing a white chip in (i,j) would be okay unless the row i is black. Wait, this is getting confusing. Let me formalize it.

Let me denote:

For each row, it can be in one of three states: white (all chips in that row are white), black (all chips in that row are black), or empty (no chips in that row).

Similarly, each column can be white, black, or empty.

However, the color assignments of rows and columns must be compatible. Specifically, if a row is colored (white or black) and a column is colored, their intersection cell must be of the same color. Therefore, if row i is white and column j is black, then cell (i,j) must be empty. Conversely, if cell (i,j) is occupied, then the color of row i and column j must both be that color, or one of them can be uncolored?

Wait, actually, if cell (i,j) has a white chip, then row i must be white or empty, and column j must be white or empty. Wait, no, the row and column must have all their chips of the same color. Therefore, if there is a white chip in row i, then the entire row must be white (i.e., all non-empty cells in the row are white). Similarly for columns. Therefore, if a row has any white chips, then the entire row is considered white, and similarly for columns.

Therefore, the rows and columns that are colored (non-empty) must form a sort of "grid" where their colors agree on intersections. So, if row i is white and column j is white, then their intersection can be white. If row i is white and column j is black, their intersection must be empty. Similarly, if a row is white and a column is empty, then the intersection can be white (since the column being empty doesn't impose a color). Wait, but if the column is empty, that means it has no chips, so the cell (i,j) could be part of the row's white chips. Hmm, perhaps the columns that are empty can have cells filled as per their row's color, but since the column is empty, it doesn't impose any color constraint. Wait, but the column being empty just means that it has no chips, so if a cell in that column is filled by its row's color, that's okay. Wait, but if a column is empty, that means none of its cells have chips, right? Wait, no, actually, the column's state is determined by its chips. If a column has any chips, they must all be the same color, so if a column has any chips, it's colored (white or black); if it has no chips, it's empty. Similarly for rows.

Therefore, the entire grid can be thought of as having some rows colored (white or black) and some columns colored (white or black), with the constraint that whenever a row and column are both colored, their intersection cell must be of the same color. However, the cells not in any colored row or column can be filled or not? Wait, no, actually, the cells in a colored row must be filled with that color, except where blocked by a column of a different color. Wait, perhaps this is getting too abstract.

Alternatively, let's model the problem as follows:

Let R be the set of rows that are colored (either white or black), and C be the set of columns that are colored (either white or black). For each row in R, it has a color (W or B), and similarly for each column in C. The coloring must satisfy that for any row r in R and column c in C, if r and c are colored, then their colors must be the same at their intersection. Wait, actually, no. If row r is white and column c is black, then their intersection cell (r,c) cannot have a chip, because that cell would need to be both white and black. Therefore, such a cell must be empty. However, if row r is white and column c is white, then the cell (r,c) can be white. If a row is white and a column is empty, then the cell (r,c) can be white (since the column has no chips, so it's allowed). Similarly, if a column is white and the row is empty, the cell can be white.

Therefore, the total filled cells are those in colored rows or columns, except where a colored row and colored column of different colors intersect (those cells are empty). However, the problem requires that the placement is maximal. So, we can't add any more chips. That would mean that any cell not already filled is blocked by either its row or column being colored with a different color. Wait, let me think.

To ensure maximality, every empty cell must be in a row that is colored with a different color than the column it's in, or something like that? Let me see:

Suppose a cell (i,j) is empty. To prevent adding a white chip there, either row i is black or column j is black. Similarly, to prevent adding a black chip there, either row i is white or column j is white. Therefore, to block both possibilities, the cell must be in a row and column that are both colored with opposite colors. That is, if row i is white and column j is black, then you can't add a white chip (because column j is black) and can't add a black chip (because row i is white). Similarly, if row i is black and column j is white, same thing. Alternatively, if the row is colored and the column is colored with the same color, then you could add a chip of that color, so that cell must already be filled. Wait, but if the row and column are both white, then their intersection is already white, so it's filled. So, the only empty cells are those where the row and column are colored with different colors. Therefore, in such a configuration, all empty cells are intersections of differently colored rows and columns, and all other cells are filled with the color of their row or column (whichever is colored, since if only the row is colored, then the cell is that color; if only the column is colored, same; if both are colored the same, then it's that color).

Therefore, such a configuration is determined by a partition of the rows and columns into colored ones (with colors) such that the intersections of differently colored rows and columns are empty, and all other cells are filled. Additionally, the configuration must be maximal, which in this case would mean that you can't add any more colored rows or columns without causing a conflict. Wait, perhaps not exactly. Wait, actually, the maximality here is about the placement of chips, not the coloring. So, the coloring is a result of the chip placement. Therefore, the maximal condition translates to the fact that you can't add any more chips without violating the row or column color constraints. Which, as we saw earlier, requires that every empty cell is at the intersection of a colored row and a colored column with opposite colors. Therefore, such a configuration is completely determined by a set of colored rows (each white or black) and colored columns (each white or black), such that:

1. For any row colored white and column colored black, their intersection is empty (which is already enforced by the coloring).

2. All other cells (those in colored rows or columns with matching colors) are filled.

3. Every empty cell is such an intersection of differently colored rows and columns.

Additionally, since we have a finite number of chips (25 white and 25 black), we need to ensure that the total number of white chips and black chips used in the configuration does not exceed 25 each. Wait, but actually, the problem says "some of these chips", so we can use any number up to 25 of each, but the placement must be maximal. However, since the problem says "place some of these chips", but the maximal condition requires that you can't add any more, so the number of chips is determined by the configuration. So, the count is over all such maximal configurations, regardless of the number of chips used, as long as they don't exceed 25 of each. Wait, but since the grid is 5x5, the maximum number of chips is 25, so using 25 of one color would mean the other color can't be used. Hmm, but the problem allows using some of the chips, so maybe the count includes all maximal configurations, even those that use fewer chips. However, the problem says "some of these chips", so maybe we can use any number, but the maximal condition requires that you can't add more. So, the configurations can vary in size, but must be maximal.

Therefore, the problem reduces to counting all possible such colorings of some rows and columns (with colors white or black) such that:

- The colored rows and columns form a "grid" where conflicting intersections are empty.

- Every empty cell is at the intersection of a colored row and a colored column with opposite colors.

Additionally, the total number of white chips (cells in white rows or columns where the column/row color matches) must be ≤25, and similarly for black. But since the grid is 5x5, the total chips in any configuration can't exceed 25, so as long as the white and black counts are each ≤25, which they will be automatically, since the total is 25. Wait, actually, if all 25 cells are filled with white, that would use 25 white chips and 0 black, which is allowed. Similarly for black. So, the constraints on the number of chips are automatically satisfied as long as the configuration is valid. Therefore, the problem is just to count all such maximal configurations.

Therefore, the problem reduces to counting the number of pairs (R, C), where R is a set of rows each assigned white or black, C is a set of columns each assigned white or black, such that:

- For any row r in R and column c in C, if r is white and c is black, or vice versa, then their intersection is empty (which is already enforced by the coloring).

- Every empty cell is such an intersection of a colored row and column with opposite colors. That is, there are no empty cells that are in an uncolored row or uncolored column. Because if a cell is in an uncolored row and an uncolored column, then you could color that row and column with any color and place a chip there, violating maximality. Wait, exactly! So, to ensure maximality, there can be no empty cells that are in uncolored rows and uncolored columns. Because if there is such a cell, you could place a chip there (choosing a color, say white), which would require that the row and column be colored white (since they were previously uncolored). Since they were uncolored, there were no conflicting colors, so this would be allowed, contradicting the maximality.

Therefore, the maximality condition requires that all empty cells are intersections of colored rows and columns with opposite colors. Therefore, there are no empty cells in uncolored rows or uncolored columns. Therefore, every uncolored row must be entirely filled (since if a row is uncolored, that means it has no chips, but then any cell in that row could be filled, so to prevent that, the entire row must be blocked by the columns. Wait, this is getting a bit tangled again.

Wait, let me rephrase:

To satisfy maximality, every empty cell must be blocked by either its row or column being colored with the opposite color. Therefore, there cannot exist an empty cell that is in an uncolored row and an uncolored column. Because such a cell could be filled with either color, which would require coloring the row and column accordingly, which was not previously blocked. Therefore, the set of empty cells must be exactly the intersections of colored rows and columns with opposite colors, and all other cells (those in uncolored rows or uncolored columns) must be filled. Wait, but if a row is uncolored, that means it has no chips, so all its cells must be empty? Wait, no. Wait, if a row is uncolored, that means it has no chips, so all its cells are empty. Similarly for columns. But then, if there's an uncolored row, then all the cells in that row are empty, but those cells could potentially be filled unless blocked by their columns. But if the column is also uncolored, then the cell is in an uncolored row and column, so you could place a chip there, which would require coloring the row and column, so that's not allowed in a maximal configuration. Therefore, in a maximal configuration, there can be no uncolored rows or columns? Wait, that can't be, because you could have some rows and columns uncolored but their cells are blocked by the other direction.

Wait, perhaps the only way to have a maximal configuration is that all rows and columns are colored. Because if there's an uncolored row, then you can choose to color it (with some color) and fill all its cells (provided the columns allow it). Wait, but if the columns intersecting that row are already colored with conflicting colors, then you can't color the row. Hmm.

Alternatively, suppose there is an uncolored row. Then, for each cell in that row, either the column is colored (so the cell is blocked by the column's color), or the column is uncolored. But if the column is uncolored, then that cell is in an uncolored row and column, so you could place a chip there, which would require coloring the row and column, so that's not allowed. Therefore, in a maximal configuration, every uncolored row must have all its columns colored with some color, so that the cells in the uncolored row are blocked by their columns. Similarly, every uncolored column must have all its rows colored with some color. Wait, this is getting complex. Let me think of it as a bipartite graph.

Let me model the rows and columns as nodes in a bipartite graph, with edges representing cells. Coloring a row or column as white or black imposes constraints on the edges. Hmm, perhaps another approach.

Let me consider that in a maximal configuration, the set of colored rows and columns must form a "blocking set" such that every empty cell is blocked by a colored row or column of the opposite color. Alternatively, the complement of the filled cells must be exactly the intersections of differently colored rows and columns.

Alternatively, perhaps the configuration is equivalent to choosing a subset of rows colored white, a subset colored black, a subset of columns colored white, and a subset colored black, such that:

- No row is both colored white and black (obviously).

- Similarly for columns.

- The white rows and black columns must not intersect (since their intersection would have to be empty, but if a cell is in a white row and black column, it must be empty, which is okay).

Wait, actually, the key is that the filled cells are those in white rows or white columns (but not both if they conflict), but this is getting too vague.

Alternatively, let's think of the grid as being partitioned into four regions:

1. Cells in white rows and white columns: these are filled white.

2. Cells in white rows and black columns: these must be empty (conflict).

3. Cells in black rows and white columns: empty.

4. Cells in black rows and black columns: filled black.

Additionally, cells in uncolored rows or columns:

Wait, if a row is uncolored, then it has no chips, so all its cells must be empty unless the column is colored. Wait, this is getting too convoluted. Maybe it's better to think in terms of the entire grid being covered by colored rows and columns, except for the conflicting intersections.

Wait, here's an idea: Let me suppose that all rows and columns are colored, either white or black. Then, the filled cells are those where the row and column colors match, and the empty cells are where they differ. However, in this case, the configuration would be maximal because any empty cell (where row and column colors differ) cannot be filled without conflicting with either the row or column color. Additionally, since all rows and columns are colored, there are no uncolored rows or columns, so there's nowhere to add a chip without conflicting. Wait, but in this case, the total number of chips would be the number of cells where row and column colors match. However, this might not use all possible chips, but since we can choose the colors of rows and columns, maybe this is a way to get maximal configurations.

However, the problem allows some rows and columns to be uncolored (i.e., have no chips), but in that case, as we saw earlier, the uncolored rows and columns must be entirely blocked by the other direction. Hmm.

Alternatively, perhaps the maximal configurations are exactly those where all rows and columns are colored, because otherwise you could color an uncolored row or column. Wait, but coloring a row requires that all its cells (except those blocked by columns) are filled, but if the row is uncolored, that means it has no chips, so all its cells are empty. If I decide to color it white, then all its cells must be white, but that would require that the columns intersecting it don't have conflicting colors. So, if I have an uncolored row, I can choose to color it white only if all the columns intersecting it are either white or uncolored. But if some columns are black, then coloring the row white would conflict at those intersections. Therefore, to color an uncolored row, you need that all columns intersecting it are either uncolored or already white. Similarly for black. Therefore, if there exists an uncolored row where some columns are white and others are black, then you can't color that row without conflict. But in a maximal configuration, you must not be able to color any uncolored row or column without conflict. Therefore, for every uncolored row, the columns intersecting it must be colored with both colors, making it impossible to color the row. Similarly for columns.

Wait, this is getting too abstract. Let me try small cases.

Suppose the grid is 1x1. Then, the problem would be: place some chips (0 or 1) such that the single cell's row and column (which is itself) have the same color, and it's maximal. So, maximal means that if you place a chip, you can't add another (but there's only one cell). So the maximal configurations are either placing a white or black chip, or not placing any? Wait, but if you don't place any, then you can add a chip, so it's not maximal. Therefore, the maximal configurations are the two options: white or black. So the answer would be 2. Let me see if that fits into the above model.

In the 1x1 case, the rows and columns can be colored or not. To have a maximal configuration:

- If you color the single row white and the single column white, then the cell is filled white. This is maximal because you can't add anything else.

- Similarly, coloring row and column black gives a black chip, which is maximal.

- If you leave the row and column uncolored, then the cell is empty, but you can add a chip, so it's not maximal.

- If you color the row white and column black, then the cell must be empty (due to conflict), but then you could add a chip? Wait, no, because the row is white and column is black, so you can't place a chip there (since it would have to be both colors). Therefore, that configuration (row white, column black) is actually allowed, but it's empty. However, since it's empty, you could have placed a chip by choosing not to color the row and column. Wait, but in this case, if the row is white and column is black, the cell is empty, but you can't add a chip there because of the conflicting colors. However, you could choose to not have the row and column colored, which would allow placing a chip. Wait, but the configuration with row white and column black is a different configuration from the uncolored one. The problem counts colorings where rows and columns are colored in such a way that the resulting chip placement is maximal. So in the 1x1 case, the configurations where row and column are colored with different colors result in an empty cell which is maximal? Because you can't add a chip there. Wait, actually, in that case, you can't add a chip, so it is maximal. However, the empty cell with uncolored row and column is not maximal because you can add a chip. Therefore, the maximal configurations are:

1. Row and column both white: filled white.

2. Row and column both black: filled black.

3. Row white and column black: empty.

4. Row black and column white: empty.

Wait, but in cases 3 and 4, the cell is empty, but you can't add a chip there because of the conflicting colors. So those are maximal. However, the problem allows using some of the chips, so the empty configuration (all rows and columns uncolored) is not maximal, but the configurations where rows and columns are colored in a way that blocks the cell are maximal even if they result in zero chips. Wait, but the problem says "place some of these chips", so does the empty configuration count? The problem says "some", which might mean at least one, but maybe not. The problem says "some" which could include zero. Hmm, but in the 1x1 case, if we consider the empty configuration with rows and columns uncolored, that's not maximal, so it's excluded. The configurations with row and column colored oppositely are maximal with zero chips. So in total, there are 2 (with one chip) plus 2 (with zero chips) = 4? But that contradicts my earlier thought. Wait, but in the problem statement, it says "place some of these chips", so maybe the empty configuration is allowed as "some" (since "some" can mean zero or more). However, the maximality condition requires that you can't add any chips. The empty configuration with all rows and columns uncolored is not maximal because you can add a chip. The configurations where rows and columns are colored to block the cell (e.g., row white and column black) are maximal with zero chips. So in the 1x1 case, there are 2 (one chip) + 2 (zero chips) = 4 maximal configurations? But that seems too many. Wait, actually, in the 1x1 case, the colorings of the row and column can be:

- Row white, column white: cell is white (1 chip).

- Row white, column black: cell empty (0 chips).

- Row black, column white: cell empty (0 chips).

- Row black, column black: cell is black (1 chip).

- Row uncolored, column uncolored: cell empty (0 chips, but not maximal).

- Row uncolored, column white: cell white (but row is uncolored, so the row has no chips, so the cell must be empty? Wait, no. If the column is white, then the cell must be white, even if the row is uncolored. Wait, the column being white requires that all chips in the column are white. Since the column has one cell, which would be filled white, so the row would have a white chip, so the row is effectively white. Wait, this is conflicting.

Hold on, maybe my initial assumption that rows and columns can be independently colored is wrong. Because if a column is colored white, then all the cells in that column must be white (if they are filled). But if the corresponding row is uncolored, does that mean the row can have a white chip? Wait, the row's color is determined by its chips. If a cell in the row is filled (because the column is white), then the row must be considered white. Therefore, you can't have a column colored white and a row uncolored if they intersect. Therefore, the coloring of rows and columns must be consistent: if a column is colored, then the rows intersecting it must either be uncolored or colored the same as the column? Wait, no, the row's color is determined by its chips. If a column is white, then the cell at the intersection with a row must be white if the row is colored white, but if the row is uncolored, then the cell can be white (since the column is white). However, if the row is uncolored but has a white chip (from the column's coloring), then the row is effectively white. Therefore, the row must be considered white in that case. Therefore, you can't have a column colored white and a row uncolored if they intersect, because the row would have a white chip, hence must be colored white. Therefore, the colorings must be such that for any column colored white, all rows intersecting it must either be white or uncolored (but if uncolored, the row would have a white chip, so actually must be white). Therefore, in reality, if a column is colored, all rows intersecting it must be colored the same color. Similarly for rows.

This suggests that the rows and columns must form a "rectangle" of color. For example, all white rows must include all columns that are white, and vice versa. Wait, perhaps the only way this works is if the set of white rows and white columns form a rectangle, and similarly for black. Wait, this is getting too tangled. Maybe I need to think of the problem as a matrix where the filled cells form a rectangle of one color and another rectangle of another color, but not overlapping?

Alternatively, let's consider that in order to have a maximal configuration, the colored rows and columns must form a partition of the grid into monochromatic rectangles, with the intersections of different colors being empty. Wait, perhaps the problem is equivalent to choosing a subset of rows and columns to color white or black such that the white rows and white columns form a rectangle of white cells, the black rows and black columns form a rectangle of black cells, and these rectangles don't overlap (since overlapping cells would need to be both colors). Additionally, the union of these rectangles must cover all cells except those blocked by conflicting colors. Hmm, maybe not.

Alternatively, here's another approach inspired by linear algebra: think of the grid as a matrix where each row and column is either all 0, all 1, or all empty (but in our case, the colors are white and black, so maybe 0 and 1). The condition is that in the final matrix, every row and column is monochromatic. The maximal condition means that you can't add any more elements without violating the monochromaticity. This is similar to a maximal monochromatic rectangle or something.

Wait, actually, in such a maximal arrangement, every row that isn't entirely filled must have at least one column that intersects it with an opposite color, preventing it from being extended. Similarly for columns.

Alternatively, the problem might be equivalent to choosing a pair of subsets (R_w, C_w) for white and (R_b, C_b) for black such that:

- R_w and C_w are sets of rows and columns colored white, with R_w × C_w being the white cells.

- Similarly, R_b and C_b are sets for black, with R_b × C_b being black cells.

- The white and black regions don't overlap (so R_w ∩ R_b = empty, and similarly for columns? Not exactly, but their intersections must be empty.

Wait, actually, the white cells are the intersection of white rows and white columns, but if a row is white and a column is black, their intersection is empty. So the white cells are R_w × (all columns not in C_b) ∩ (all rows not in R_b) × C_w? Hmm, this is getting too convoluted.

Alternatively, let me think of the grid as being divided into four regions:

1. Rows in R_w (white rows) and columns in C_w (white columns): these cells are white.

2. Rows in R_w and columns in C_b (black columns): these cells must be empty (since row is white and column is black).

3. Rows in R_b (black rows) and columns in C_w: empty.

4. Rows in R_b and columns in C_b: black cells.

Additionally, rows not in R_w or R_b and columns not in C_w or C_b can have cells that are uncolored, but in a maximal configuration, these must be blocked by the existing colors. Wait, but in a maximal configuration, there can be no such rows or columns, because otherwise you could color them.

Wait, perhaps the maximal configurations are those where every row is either in R_w or R_b, and every column is in C_w or C_b. Because if there's a row not in R_w or R_b, then you could color it (provided the columns allow it). To prevent that, all rows must be colored (either white or black), and similarly for columns. Therefore, in a maximal configuration, all rows and columns must be colored (either white or black), and the coloring must be such that no two colored rows and columns of different colors intersect (except in empty cells). Wait, but they do intersect in empty cells.

Therefore, the problem reduces to counting the number of 2-colorings (white/black) of all rows and all columns such that:

- For any row colored white and column colored black, their intersection is empty (which is enforced by the coloring).

- The coloring is such that you can't recolor any row or column to add more chips. Wait, but since all rows and columns are already colored, you can't add any more chips because any empty cell is at the intersection of differently colored rows and columns. Therefore, such a configuration is maximal.

Therefore, the problem reduces to counting the number of pairs ( (R_w, R_b), (C_w, C_b) ), where R_w is the set of white rows, R_b the set of black rows (disjoint, covering all rows), and similarly C_w and C_b for columns, with the condition that:

- The white rows and white columns form a rectangle of white cells (their intersection),

- The black rows and black columns form a rectangle of black cells,

- The white and black rectangles do not overlap (since their intersection would require a cell to be both colors).

Additionally, the white and black regions must cover all cells except those in conflicting row/column color intersections. Wait, actually, the total filled cells are (number of white rows × number of white columns) + (number of black rows × number of black columns). The empty cells are (white rows × black columns) + (black rows × white columns).

Since the grid is 5x5, we have:

Let me denote:

Let W_r = number of white rows,

B_r = number of black rows = 5 - W_r,

W_c = number of white columns,

B_c = number of black columns = 5 - W_c.

Then the total white cells are W_r * W_c,

Total black cells are B_r * B_c,

Total empty cells are W_r * B_c + B_r * W_c.

The total must satisfy:

White cells ≤25,

Black cells ≤25,

And the arrangement must be such that all rows and columns are colored (so no uncolored rows or columns). Since we are considering maximal configurations where all rows and columns are colored, this covers all possibilities.

Therefore, the problem reduces to counting all possible choices of W_r (from 0 to 5), W_c (from 0 to 5), and then assigning the remaining rows and columns as black, such that:

- The white cells (W_r * W_c) ≤25,

- The black cells ((5 - W_r)*(5 - W_c)) ≤25,

But since W_r and W_c can range from 0 to 5, these inequalities are always satisfied except when W_r=5 and W_c=5 (which would give 25 white cells and 0 black, which is okay), or similar.

However, we also need to consider that the coloring must be consistent. That is, once you choose W_r white rows and W_c white columns, the remaining rows are black, and the remaining columns are black. Therefore, for each possible W_r (0-5) and W_c (0-5), there is exactly one way to assign the colors to rows and columns to get that configuration. However, we also have to consider that the assignment of which specific rows and columns are white or black matters. Wait, actually, the count is over all possible subsets of rows and columns colored white or black, so for each W_r and W_c, the number of configurations is C(5, W_r) * C(5, W_c). However, we have to consider that the color assignments must be such that the rows not chosen as white are black, and similarly for columns. Therefore, for each W_r and W_c, the number of configurations is indeed C(5, W_r) * C(5, W_c). However, we must also ensure that the white and black regions don't overlap, which they automatically don't because the black rows and columns are the complements.

Wait, but actually, the overlapping would be between white and black regions? No, the white region is W_r × W_c, and the black region is B_r × B_c = (5 - W_r) × (5 - W_c). These two regions don't overlap because their rows and columns are disjoint. The overlapping of white and black would be if a row is both white and black, which isn't allowed, but since we partition the rows into white and black, that's okay.

Therefore, the total number of such configurations is the sum over W_r from 0 to 5 and W_c from 0 to 5 of [C(5, W_r) * C(5, W_c)]. However, this counts all possible such colorings, but we need to exclude those configurations where the white or black regions exceed 25 chips? Wait, no, since the maximum white cells is 5*5=25 when W_r=5 and W_c=5, which is allowed. Similarly for black. So all configurations are allowed as long as they are maximal, which in this case they are because all rows and columns are colored, so you can't add any more chips.

Wait, but hold on. The problem states that we have 25 white chips and 25 black chips available. However, in the configurations we are counting, the number of white chips is W_r * W_c and black chips is (5 - W_r)*(5 - W_c). Since we can't use more than 25 of each, but since W_r and W_c are at most 5, the maximum white chips is 25 (when W_r=5, W_c=5), and similarly for black. So all these configurations are valid in terms of chip counts.

However, the problem says "place some of these chips", so using fewer is okay. Therefore, the total number of configurations is indeed the sum over all W_r and W_c of C(5, W_r)*C(5, W_c). But wait, that would be (sum_{W_r=0 to 5} C(5, W_r)) * (sum_{W_c=0 to 5} C(5, W_c)) ) = (2^5)*(2^5) = 1024. But this counts all possible ways of choosing white rows and white columns, with the rest being black. However, this includes configurations where, for example, W_r=0 and W_c=0, which would mean all rows and columns are black. Wait, no, if W_r=0, then all rows are black, and W_c=0 means all columns are black, so the entire grid is black (25 black chips). Similarly, W_r=5 and W_c=5 would be all white. But also, configurations like W_r=1, W_c=1 would have 1 white cell and 16 black cells (since (5-1)*(5-1)=16). Wait, but the problem requires that the placement is maximal. However, in our earlier analysis, we considered that maximal configurations require all rows and columns to be colored, which is exactly what this count is doing. Therefore, all these configurations are maximal because you can't add any more chips. For example, if you have W_r=1, W_c=1, then the single white cell is at the intersection of the white row and column. All other cells in the white row are in black columns, so they can't be white, and all other cells in the white column are in black rows, so they can't be white. The remaining cells are in black rows and columns, so they can be black, but they already are. Wait, actually, in this case, the black region is (5-1)*(5-1)=16 cells, so all those are filled black. Therefore, the entire grid is filled except for the intersections of white row with black columns and black rows with white columns, which are empty. Therefore, you can't add any more chips because any empty cell is at the intersection of a white and black row/column, so you can't place a chip there. Therefore, all these configurations are indeed maximal.

Therefore, the total number is 2^5 * 2^5 = 1024. But wait, the problem specifies that the chips are placed "in the 25 unit cells", so the empty cells are allowed as long as the conditions hold. However, the problem might be considering two configurations different if the set of white rows or columns differ, even if the resulting chip placement is the same. Wait, but in our model, different choices of white rows and columns lead to different chip placements. For example, choosing different white rows but same number would lead to different white cells. Therefore, the count of 1024 is correct?

Wait, but hold on. Let me think of a small case again. Take 1x1 grid. Then according to this formula, it would be (2^1)*(2^1)=4, which matches our earlier analysis where there are four configurations: white, black, white row and black column (empty), black row and white column (empty). However, in the problem statement, the chips are indistinct except for color, so the empty configurations where you have conflicting rows and columns are considered different based on the row and column colors. But in terms of the actual chip placement (i.e., the grid's appearance), the two empty configurations (white row & black column vs. black row & white column) would look the same (both empty), but since the rows and columns are colored differently, are they considered different configurations?

The problem says "the number of ways to place some of these chips...", so the placement is determined by the positions and colors of the chips. The coloring of the rows and columns is a result of the placement, not an independent choice. Therefore, two configurations that result in the same chip colors in the same cells are the same, even if the row/column color assignments differ. Wait, but actually, the row and column colors are determined by the chip placements. For example, if you have an empty cell because the row is white and column is black, versus the row is black and column is white, the actual grid is the same (empty), but the row and column colors are different. However, since the problem doesn't label the rows and columns, only the grid cells, perhaps these are considered the same?

Wait, the problem says "the number of ways to place some of these chips in the 25 unit cells...", so the answer is based on the grid's cell configurations, not the row/column colorings. Therefore, two different row/column colorings that result in the same grid (same cells with same colors) are considered the same configuration. Therefore, we need to count distinct grid colorings, not the row/column assignments.

Oh! This is a crucial point I missed earlier. The problem is asking for the number of distinct grid arrangements (i.e., assignments of white, black, or empty to each cell) that satisfy the conditions. The row and column colorings are a tool to construct such arrangements, but different row/column assignments might lead to the same grid.

Therefore, my previous approach of counting row and column choices overcounts because different row/column colorings can lead to the same grid.

For example, in the 1x1 case, choosing row white and column black gives an empty cell, as does choosing row black and column white. But both result in the same grid (empty), so they should be counted as one, not two. However, the problem might consider the empty grid as a valid configuration (if it's maximal). Wait, in the 1x1 case, the empty grid with no chips is not maximal because you can add a chip. The two configurations where the cell is empty but rows/columns are colored oppositely are maximal, but they both result in the same grid (empty cell), but with different row/column colorings. However, since the problem is about the grid's state, not the row/column colors, those two would be considered the same. Therefore, the count should be 3 for 1x1: white, black, or empty (but empty is not maximal). Wait, no, the empty configuration with rows and columns uncolored is not maximal, but the empty configuration with rows and columns colored oppositely is maximal. However, the grid's appearance is the same (empty cell). So, does the problem count them as different?

The problem says "the number of ways to place some of these chips...", so the "way" is determined by the placement of chips (their positions and colors). The row and column colorings are just a method to ensure the constraints. Therefore, two different row/column colorings that result in the same chip placement are the same "way".

Therefore, in the 1x1 case, the maximal configurations are:

1. White chip.

2. Black chip.

3. Empty cell with row and column colored oppositely (but this is just an empty cell, same as another such configuration).

Wait, but the empty cell can be achieved in two different ways (row white/column black or row black/column white), but the grid is the same. Therefore, the problem would count this as one configuration (the empty grid), but it's maximal? Wait, but the empty grid (with all rows and columns uncolored) is not maximal, but the empty grid with rows and columns colored oppositely is maximal. However, the grid's appearance is the same (empty). So, does the problem consider these as the same configuration?

This is ambiguous, but I think the problem considers two configurations different only if the set of chips (their positions and colors) differ. Therefore, the row/column colorings are just a tool, and the actual count is based on the grid's visible state. Therefore, in the 1x1 case, there are two maximal configurations: white or black. The empty grid is not maximal, so it's excluded. The two different row/column colorings that lead to an empty cell are not considered because the empty cell isn't maximal. Wait, but if you have row white and column black, the cell is empty and maximal. But the problem might consider that as a valid configuration even though it's empty, as long as it's maximal. So in that case, the empty cell with the row and column colored oppositely is a valid maximal configuration, but it's just an empty grid. However, the problem says "place some of these chips", which might allow zero chips. So, in that case, the 1x1 grid would have three maximal configurations: white, black, and empty (with conflicting rows/columns). But the empty one is different from the uncolored empty grid because it's blocked. However, visually, they both appear empty. Therefore, the problem might count them as the same, so only two configurations. This is ambiguous.

Given the problem is from a competition, likely the answer is 1024, but that seems too high. Wait, but in the problem statement, the chips are placed such that all chips in the same row and column have the same color. Therefore, if a row has any chips, they must all be the same color. Similarly for columns. Therefore, the only way to have an empty cell is if its row and column have different colors. However, the empty grid (no chips) trivially satisfies the conditions (since there are no chips in any row or column, so the color condition is vacuously true). But it's not maximal because you can add a chip. Therefore, the empty grid is excluded. The configurations where some rows and columns are colored such that their intersections are empty are considered different based on their row and column colorings, but the problem counts the grid's chip placements. Therefore, two different row/column colorings that result in the same chip placement are the same.

Therefore, the correct count is the number of possible assignments of colors to rows and columns (with all rows and columns colored) divided by the equivalence of row/column permutations? No, not exactly. Wait, actually, each choice of white rows and white columns gives a unique grid pattern. For example, choosing different sets of white rows and columns will result in different white and black regions. Therefore, the total number is indeed the number of ways to choose white rows and white columns, which is 2^5 * 2^5 = 1024. However, this includes configurations where all rows and columns are black (W_r=0, W_c=0), which gives a full black grid (25 black chips), and similarly all white. It also includes configurations where, say, W_r=1 and W_c=0, which would mean one white row and all columns black. The white row would have its cells empty because all columns are black except none (since W_c=0). Wait, no, if W_c=0, then all columns are black. So a white row and black columns would have their intersections empty. Therefore, the white row has no chips (since all columns are black, conflicting with the white row). Therefore, the white row is actually empty, so it shouldn't be considered a white row. Wait, this is a contradiction.

Ah, here's the problem with my earlier reasoning. If I choose W_r white rows and W_c white columns, but if a white row has all its columns colored black, then that row actually can't have any chips, so it shouldn't be considered a white row. Therefore, my initial assumption that you can independently choose any subset of rows and columns as white is flawed because the intersection must be non-empty? Wait, no, the rows and columns can be colored even if their intersection is empty. For example, you can have a white row with no white columns, resulting in an empty row. But according to the problem's conditions, a row that is colored white must have all its chips white, but if all columns are black, then the row can't have any chips, so it's allowed. However, the row is still considered white (since you chose it as such), even though it has no chips. Similarly for columns.

However, the problem requires that the placement is maximal. If you have a white row with no chips (because all columns are black), then you could potentially change the column colors to allow some white chips. Wait, but in the maximal configuration, you can't add any chips. If the row is white and a column is black, their intersection is empty. To add a white chip there, you would have to recolor the column as white, but that might affect other rows. However, the maximality condition is about adding a chip without changing existing colors. Wait, the problem says "any additional chip placed on the grid would violate one or more of the previous two conditions".

Therefore, adding a chip requires that the row and column already have chips of the same color. If a row is white and a column is black, you can't place a white chip because the column is black, and can't place a black chip because the row is white. Therefore, such a cell is blocked. Therefore, even if a white row has all columns black, making it empty, it is still a valid part of a maximal configuration. The row is white, the columns are black, so you can't add anything there. Therefore, such configurations are allowed.

Therefore, the initial count of 2^5 * 2^5 = 1024 is correct, but considering that each choice of white rows and white columns gives a distinct configuration (since the positions of white and black chips depend on the intersection). For example, choosing different white rows or columns will lead to different white and black regions on the grid. Therefore, each combination of white rows and columns gives a unique grid configuration. Therefore, the answer is 1024. However, the problem specifies that we have 25 white and 25 black chips available, but since the configurations can use up to 25 of each, and the count includes all such possibilities, this should be okay.

Wait, but in the problem statement, it says "some of these chips", so using zero chips is allowed as long as it's maximal. The configuration with all rows and columns colored such that every cell is blocked (e.g., all rows white and all columns black) would result in zero chips and is maximal. However, such a configuration is counted in the 1024. But the problem might require that at least one chip is placed. If so, we need to subtract the all-empty configuration. But how many all-empty configurations are there? Only one, where all rows and columns are colored such that every cell is blocked. Wait, actually, to have all cells empty, you need that for every cell (i,j), row i and column j are colored differently. This requires that the set of white rows and black columns form a sort of covering. Actually, the all-empty configuration can be achieved in multiple ways. For example, choosing all rows white and all columns black, or vice versa, or other combinations where every row and column have opposite colors. Wait, but to have all cells empty, for every row and column, if a row is white, all columns must be black, and vice versa. Therefore, the only way to have all cells empty is to have all rows one color and all columns the other. So either all rows white and all columns black, or all rows black and all columns white. These are two distinct configurations in the count (since they choose W_r=5, W_c=0 or W_r=0, W_c=5). Therefore, if the problem requires at least one chip, we need to subtract these two configurations. However, the problem says "some of these chips", which could include zero. So unless specified otherwise, we should include them.

But let me check the problem statement again:

"Find the number of ways to place some of these chips in the 25 unit cells of a 5×5 grid such that:

each cell contains at most one chip

all chips in the same row and all chips in the same column have the same colour

any additional chip placed on the grid would violate one or more of the previous two conditions."

The term "some" can include zero, but the third condition requires maximality. The all-empty configuration with all rows and columns colored oppositely is maximal (you can't add any chips), so it should be counted. The all-empty grid with no rows or columns colored is not maximal, so it's excluded. Therefore, the two configurations (all rows white/columns black and vice versa) are valid and should be included. Therefore, the total is indeed 1024.

But wait, in the problem statement, the chips are indistinct except for color. So, the count is over colorings of the grid where each cell is white, black, or empty, satisfying the conditions. Each choice of white rows and white columns gives a unique such coloring. Therefore, the total number is 2^5 * 2^5 = 1024. However, this seems too large for an Olympiad problem, which usually has more manageable numbers. Maybe I made a mistake.

Wait, another way to think: For each row, you can choose to make it white, black, or empty. Similarly for each column. But with the constraints that:

- If a row is non-empty (white or black), then all its non-empty cells must be of that color.

- Similarly for columns.

- The arrangement is maximal.

But this approach might not be straightforward.

Alternatively, the problem might be equivalent to choosing a rectangle of white chips and a rectangle of black chips such that they don't overlap, and together they block all other cells. The maximal condition requires that the entire grid is covered by the white rectangle, black rectangle, and their blocking regions. Therefore, the white and black rectangles must be as large as possible.

Wait, perhaps the configurations are determined by choosing a rectangle for white and a rectangle for black such that they don't overlap, and their "shadows" (rows and columns) block the rest. The maximal condition would require that you can't add any more to either rectangle without overlapping.

Alternatively, the problem might be equivalent to selecting a pair of complementary rectangles covering the grid such that their colors block the rest. This is similar to the row and column coloring approach.

Alternatively, the number of maximal configurations is the number of ways to choose a subset of rows and columns for white, and the rest are black, which is indeed 2^5 * 2^5 = 1024. However, in the problem statement, the chips are placed on the grid, so the actual number of chips is variable, but the count is over all such maximal arrangements. Since the problem allows using any number of chips (as long as maximal), this would be 1024. But I suspect the answer is 96, or something similar. Wait, maybe I'm overcounting because some configurations are equivalent under row and column permutations, but the problem doesn't mention considering symmetries, so likely not.

Alternatively, perhaps the problem requires that every row and column that is colored must have at least one chip. Because otherwise, if you have a white row with no chips (because all columns are black), then that row could be removed (turned to empty) and the configuration would still be valid, contradicting maximality. Wait, this is a key point!

Ah! Here's a mistake in my previous reasoning. The maximality condition requires that you cannot add any more chips, but it also implies that you cannot remove any existing chips without violating the conditions. Wait, no, the problem doesn't say that. Maximality here is only about not being able to add more chips, not about minimality. However, if a row is colored white but has no chips (because all columns are black), then that row is "redundant" in the sense that you could uncolor it (turn it to empty) and still have a valid configuration. However, the problem doesn't require the configuration to be minimal, only maximal. Therefore, such configurations are still allowed as maximal, because you can't add chips, even if you could remove some.

However, the problem might implicitly require that all colored rows and columns actually have chips. Because if a row is colored white but has no chips, then the row's color is irrelevant, and the configuration could be simplified by uncoloring the row, but since the problem only requires maximality regarding adding chips, not removing, it's still valid.

But let's think: if a row is colored white but has no chips (all columns are black), then that row is effectively empty. However, the coloring is part of the configuration's constraints. The problem counts different colorings as different even if they result in the same chip placement but different row/column colors. Wait, but earlier we concluded that the problem counts based on the grid's appearance. Therefore, a white row with no chips (because columns are black) would have the same grid appearance as an uncolored row with columns black. But since the row is colored white, it's a different configuration in terms of row/column assignments but same in grid appearance. Therefore, the problem's answer is indeed 1024 if considering row/column assignments, but if considering grid appearances, it's less.

This is the crux of the issue. The problem says "the number of ways to place some of these chips...", so the "way" is determined by the placement (positions and colors of chips), not the row/column colorings. Therefore, two different row/column assignments that result in the same chip colors and positions are the same "way".

Therefore, we need to count the number of distinct grid colorings (each cell is W, B, or empty) that satisfy:

1. Every non-empty row is monochromatic.

2. Every non-empty column is monochromatic.

3. The placement is maximal (can't add any chip without violating 1 or 2).

Therefore, to count these, we need to consider the grid colorings where:

- All non-empty rows and columns are monochromatic.

- Every empty cell is in a row and column of opposite colors (or both empty? No, because maximality requires that you can't add a chip there, so at least one must be colored oppositely).

Wait, more precisely, for every empty cell (i,j):

- If you try to place white there, then row i must be black or column j must be black.

- If you try to place black there, then row i must be white or column j must be white.

Therefore, to block both, either:

- Row i is white and column j is black, or vice versa.

Or,

- One of them is colored and the other is uncolored but with conflicting color.

Wait, but if a row is uncolored, that means it has no chips, so the cell could be filled if the column allows. Wait, this is getting too tangled again.

Alternatively, the maximal condition requires that the only empty cells are those that are in the intersection of a white row and black column or vice versa. All other cells must be filled with the color of their row or column (if colored). Additionally, any row or column that is colored must have at least one chip (otherwise, you could uncolor it and still be maximal? Not necessarily, but it's allowed as per the problem's conditions).

Therefore, the grid must be partitioned into four regions:

1. White region: cells in white rows and white columns.

2. Black region: cells in black rows and black columns.

3. Empty regions: cells in white rows and black columns, or black rows and white columns.

Additionally, the white rows must form a set of rows, black rows the complement, white columns a set of columns, black columns the complement.

The key point is that the white and black regions must be rectangles covering parts of the grid, and their intersections with opposite colors are empty.

Therefore, the number of such grid colorings is the number of ways to choose a white rectangle (W_r rows and W_c columns) and a black rectangle (B_r rows and B_c columns) such that:

- The white and black rectangles don't overlap (i.e., their row and column sets are disjoint? Not exactly, but their intersection would be in the empty regions).

Wait, actually, the white rectangle is W_r × W_c, and the black rectangle is (5 - W_r) × (5 - W_c). These two rectangles don't overlap because their rows and columns are disjoint. The remaining cells are the empty regions.

Therefore, each such choice of W_r and W_c (from 0 to 5) gives a unique grid coloring. However, different W_r and W_c can lead to the same grid if the white and black regions are the same. Wait, no, because the white region is exactly the intersection of the chosen white rows and columns, so different W_r or W_c will lead to different shapes.

Therefore, the total number of distinct grid colorings is indeed the number of choices for W_r and W_c, multiplied by the number of ways to choose which rows and columns are white. Wait, no, the grid coloring is uniquely determined by the choice of white rows and white columns. For example, choosing rows 1-2 as white and columns 1-3 as white gives a specific white rectangle and black rectangle. Another choice of different rows and columns would give a different grid. Therefore, the total number is indeed the number of ways to choose any subset of rows and columns to be white (with the rest black), which is 2^5 * 2^5 = 1024. However, this counts all possible such configurations, including those where the white or black regions are empty.

But the problem might consider the color assignments of rows and columns as part of the configuration, even if they don't affect the grid's appearance. For example, if all rows are black and all columns are black, that's one configuration. If all rows are white and all columns are white, that's another. These are different and both valid.

However, in the problem statement, the chips are indistinct except for color, so the actual count is over the colorings of the grid cells. Each grid cell can be white, black, or empty, but subject to the conditions. The row and column choices are a way to generate these colorings, but different row/column choices can lead to the same grid.

Wait, for example, suppose I choose white rows {1} and white columns {1}. This gives a white cell at (1,1), and the rest of row 1 and column 1 are empty (since they're in black columns/rows). The black region is the remaining 4x4 grid. Alternatively, if I choose white rows {2} and white columns {2}, this gives a white cell at (2,2) and a different black region. These are different grid colorings, so they should be counted separately. Therefore, each choice of white rows and columns gives a unique grid coloring. Therefore, the total is indeed 1024.

But the problem specifies "some of these chips", so using all 25 white chips (by choosing all rows and columns as white) is allowed, as is using all 25 black. The count includes all these possibilities.

However, 1024 seems very large for an Olympiad problem. Maybe I made a mistake in assuming that all row and column colorings are allowed. Let me think again.

Wait, the problem requires that all chips in the same row have the same color. Therefore, if a row has any chips, they must all be the same color. Similarly for columns. Therefore, the only way to have a row with mixed colors is to have it empty. Wait, no, if a row has chips, they must be the same color. So the row can be empty, or all white, or all black. The same for columns.

Therefore, the grid must be partitioned into at most four regions:

1. White region: intersection of white rows and white columns.

2. Black region: intersection of black rows and black columns.

3. Empty regions: intersections of white rows with black columns and vice versa.

Additionally, the white rows and black rows partition the rows, and similarly for columns.

Therefore, each such configuration is uniquely determined by the choice of white rows and white columns. Therefore, the number is indeed 2^5 * 2^5 = 1024. However, this includes cases where the white or black regions are empty. For instance, if no rows or columns are chosen as white, then the entire grid is black (if all rows and columns are black). Similarly, choosing all rows and columns as white gives all white.

But the problem might have intended that both colors are used, but the problem statement doesn't say that. It says "some of these chips", which could mean using any number, including all white or all black.

Alternatively, perhaps the problem requires that the configuration is maximal in terms of both colors. That is, you can't add any more chips of either color. In that case, the configurations where all rows and columns are one color (say white) would not be maximal because you could add black chips in the remaining cells (but those cells are in white rows, so you can't). Wait, if all rows and columns are white, then all cells are white, so you can't add any more chips. Similarly for all black. So those are maximal.

Therefore, I think the answer is indeed 1024. However, given that the problem is from a competition, perhaps there's a different approach.

Wait another thought: The problem might require that every row and every column that is not entirely filled must be empty. That is, the only allowed rows are entirely white, entirely black, or entirely empty. Similarly for columns. Then, the maximal condition requires that you can't add a chip to any empty row or column. Therefore, every empty row must have at least one column that is colored oppositely, and every empty column must have at least one row colored oppositely.

Wait, this is similar to a bipartite graph where rows and columns are nodes, and an edge exists if the cell is filled. The maximal condition requires that the bipartite graph is such that every non-selected node (empty row or column) is adjacent to at least one node of the opposite color.

Alternatively, the problem might be equivalent to choosing a biclique (complete bipartite subgraph) in the row-column bipartite graph, with edges colored either white or black, such that the bicliques for white and black don't overlap. But I'm not sure.

Alternatively, the number of maximal configurations is equal to the number of antichains in the row and column poset, but this is too abstract.

Alternatively, let's think of the problem as each maximal configuration corresponds to a pair of subsets (R, C) where R is the set of rows with chips and C the set of columns with chips, along with a color for each subset. The color of the rows and columns must be consistent. Wait, but the colors can be different for rows and columns.

Alternatively, here's a different approach inspired by linear algebra:

Each row can be in one of three states: all white, all black, or empty.

Each column can also be in one of three states.

However, the states must be consistent: if a row is white and a column is black, their intersection must be empty.

The maximal condition requires that any empty cell is blocked by either its row or column being the opposite color.

Therefore, for each cell, if it's empty, then either its row is colored oppositely to its column, or one of them is colored and the other is empty but with conflicting color.

Wait, this is getting too involved. Given the time I've spent and the earlier reasoning leading to 1024, but suspecting that might be too high, perhaps the correct answer is 96. Wait, let me think differently.

Suppose that the maximal configuration must have all rows and columns either fully white, fully black, or fully empty, and that the arrangement is such that no two orthogonal non-empty sets have conflicting colors.

Wait, perhaps the problem is equivalent to choosing a rectangle of white chips and a rectangle of black chips such that they don't overlap and their rows and columns block the rest. The number of such configurations would be the number of ways to choose a white rectangle and a black rectangle that don't overlap.

The number of rectangles in a 5x5 grid is known, but choosing two non-overlapping rectangles of different colors would be complex.

Alternatively, the maximal configurations are those where the white and black regions form a partition of the grid into two rectangles with no overlapping rows or columns. So, choosing a subset of rows and columns for white, and the remaining rows and columns for black. This is exactly the earlier model, leading to 2^5 * 2^5 = 1024.

Alternatively, perhaps the problem requires that both colors are used, so subtract the all-white and all-black cases. Then 1024 - 2 = 1022, but this is still large.

Alternatively, maybe the problem counts the number of such configurations up to row and column permutations, but the problem doesn't specify that.

Alternatively, perhaps I made a mistake in assuming that rows and columns can be independently chosen. Suppose that if you choose W_r white rows and W_c white columns, then the black rows must be the remaining 5 - W_r rows, and black columns the remaining 5 - W_c columns. Therefore, the black region is automatically determined as the intersection of black rows and columns. Therefore, the number of configurations is indeed the number of ways to choose white rows and white columns, which is 2^5 * 2^5 = 1024.

Given that the problem is presented in a mathematical competition style, and 1024 is 2^10, which is a clean number, I might be overcomplicating it. Perhaps the answer is 1024. However, let me check small cases again.

Take 2x2 grid. Then according to this logic, the number would be 2^2 * 2^2 = 16.

Let me enumerate them:

Each configuration is determined by choosing which rows are white (0,1, or 2) and which columns are white.

For example:

- 0 white rows, 0 white columns: all black (4 black chips).

- 0 white rows, 1 white column: The white column has its cells empty because all rows are black. The black region is 2x2 minus the white column, so 2x1 black chips? Wait no, if columns are white but rows are black, the intersection cells are empty. So the black region is the intersection of black rows and black columns. If 0 white rows (all black rows) and 1 white column, then black columns are 1. So black region is 2 rows * 1 column = 2 black chips. The white column has its cells empty. So total chips: 2 black.

Similarly, 1 white row and 0 white columns: 1 white row with 0 white columns means the white row has no chips (since columns are all black). The black region is 1 black row * 2 black columns = 2 black chips. Total chips: 2 black.

Continuing, 1 white row and 1 white column: the white region is 1x1, black region is (2-1)*(2-1)=1, and empty cells are 1 (white row and black column) + 1 (black row and white column). Total chips: 1 white + 1 black.

Etc. All 16 configurations are valid, so the count holds for 2x2. Therefore, likely correct for 5x5.

Therefore, the answer is 2^5 * 2^5 = 1024. But since the problem mentions "indistinguishable" chips, but the count is over distinct arrangements, which are distinguishable by their positions and colors, so 1024 is correct.

Wait, but the problem says "indistinguishable white chips and indistinguishable black chips". So the count is over the number of distinct color patterns, which is exactly what we're counting. Each choice of white rows and columns gives a unique pattern. Therefore, the answer is 1024.

However, the problem might have a different interpretation where the rows and columns must be entirely filled if colored. That is, if you color a row white, you must fill all its cells with white (unless blocked by columns). But in that case, the only way to have a maximal configuration is to have all rows and columns colored, with the intersections of opposite colors being empty. This is the same as before, so still 1024.

Alternatively, maybe the problem requires that every colored row or column must be completely filled with chips. That is, if a row is colored white, then all its cells must be white (so the columns must also be white in those cells). This would require that the white rows and white columns form a rectangle covering all their intersections, and similarly for black. In this case, the white and black regions must be rectangles that don't overlap, and their combined rows and columns cover all rows and columns. This is the same as before, so still 1024.

Therefore, after careful consideration, I believe the answer is 1024. However, since the problem is from a competition, and 1024 is 2^10, maybe there's a different approach leading to a different answer. Alternatively, perhaps I made a mistake in considering that all row and column colorings are allowed, but actually, the colors of rows and columns must be consistent.

Wait, another thought: The color of a row and the color of a column must agree on their intersection if both are colored. Therefore, if you choose some rows to be white and some columns to be white, their intersection must be white. Similarly for black. However, if a row is white and a column is black, their intersection is empty. Therefore, the coloring is consistent as long as you don't have a row and column of different colors intersecting unless you leave that cell empty. Therefore, the count remains 2^5 * 2^5.

Alternatively, the problem might require that all non-empty rows and columns must be monochromatic, but you can have rows and columns that are partially filled. Wait, no, the problem states "all chips in the same row and all chips in the same column have the same colour", which requires that if a row has any chips, they must all be the same color. So partial filling is allowed as long as they are the same color, but the maximal condition requires that you can't add more.

Wait, but if a row is partially filled with white chips, then the columns of those chips must also be white. Therefore, the entire row could be filled white, but if some columns are black, then those cells are blocked. Therefore, the only way to have a partially filled row is if some columns are colored oppositely. However, the maximal condition requires that every empty cell in the row is blocked by a column of opposite color. Therefore, the row's color is determined, and the columns must be colored accordingly.

This again leads to the same model where rows and columns are colored, and the filled cells are the intersections of same-colored rows and columns.

Therefore, I think the answer is indeed 1024. However, since the problem specifies "indistinguishable" chips, but the arrangements are distinguishable by their positions and colors, the count is correct.

But wait, the problem says "any additional chip placed on the grid would violate one or more of the previous two conditions". This means that for any empty cell, you cannot place a chip of either color there. Therefore, for every empty cell (i,j):

- You can't place white because row i is black or column j is black.

- You can't place black because row i is white or column j is white.

Therefore, for every empty cell, either:

- Row i is white and column j is black,

- Row i is black and column j is white,

- Row i is white and column j is white but the cell is empty (but then you could place a white chip, so this is not allowed),

Wait, no. If row i is white and column j is white, then the cell must be white. So it can't be empty. Therefore, all empty cells must be at the intersection of differently colored rows and columns.

Therefore, the only empty cells are those where the row and column are differently colored. Therefore, the grid is partitioned into:

- White region: same-colored rows and columns.

- Black region: opposite-colored rows and columns' intersection? No, the black region is the intersection of black rows and columns.

Wait, the black region is the intersection of black rows and black columns. The empty regions are white rows with black columns and vice versa.

Therefore, the entire grid is covered by white region, black region, and empty regions (which are the cross between different colors).

Therefore, the count is indeed the number of ways to choose white rows and white columns, which is 2^5 * 2^5 = 1024.

Therefore, I think the answer is 1024. However, since the problem might have a different intended answer, perhaps I missed a constraint.

Wait, the problem says "each cell contains at most one chip". So empty cells are allowed, but the other conditions must hold. The maximal condition is crucial here.

Alternatively, perhaps the problem requires that every row and column that is not entirely filled must be empty. That is, you can't have a partially filled row or column. In that case, the only allowed rows are entirely white, entirely black, or entirely empty. Similarly for columns. Then, the maximal condition requires that any empty row or column is blocked by the other direction.

For example, if a row is empty, then for every column in that row, the column must be colored with some color, preventing the row from being filled. Similarly for columns.

In this case, the configurations are determined by choosing a subset of rows to be white or black, and a subset of columns to be white or black, such that:

- If a row is white, all its cells are white (so the columns must allow it).

- Similarly for black.

- Every empty row must have all columns colored (so you can't add a chip to the row).

- Every empty column must have all rows colored.

This is similar to the earlier model, leading to the same count of 1024.

Therefore, after thorough analysis, I believe the answer is 1024. However, since the problem is presented in a competition style, and 1024 is a power of two, I'll go with that.

**Final Answer**
\boxed{1024}
</think>

To solve the problem of finding the number of ways to place chips on a 5x5 grid such that each row and column has chips of the same color and the placement is maximal, we need to consider the following key points:

1. **Row and Column Coloring**: Each row and column can be colored white, black, or left empty. However, if a row or column is colored, all chips in that row or column must be of the same color.
2. **Maximality Condition**: The placement must be maximal, meaning no additional chip can be placed without violating the row or column color constraints. This implies that every empty cell must be blocked by either its row or column being colored with the opposite color.
3. **Intersection Constraints**: If a row is colored white and a column is colored black, their intersection must be empty. Similarly, if a row is colored black and a column is colored white, their intersection must also be empty.

The solution involves considering all possible ways to color the rows and columns such that the resulting grid configuration is maximal. Each row and column can independently be chosen to be white, black, or empty, but the colors must be consistent with the constraints.

Key steps:
- Each row can be independently chosen to be white, black, or empty.
- Each column can be independently chosen to be white, black, or empty.
- The intersection of a white row and black column (or vice versa) must be empty.
- The configuration must be maximal, meaning no empty cell can be filled without violating the color constraints.

The number of ways to choose the colors for the rows and columns is \(2^5\) for the rows and \(2^5\) for the columns (since each row or column can either be white, black, or empty, but the empty state is implicitly handled by the constraints). Therefore, the total number of configurations is:

\[
2^5 \times 2^5 = 32 \times 32 = 1024
\]

Thus, the number of ways to place the chips is \(\boxed{1024}\).