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 constraints must be satisfied. But the key is that the configuration is maximal, so you can't add any more chips. Therefore, the number of chips is fixed by the configuration.

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

- The colored rows and columns form a "grid" where intersections of differently colored rows and columns are empty, and all other cells are filled with the respective color.

- The coloring is maximal, meaning that you can't add another colored row or column without conflicting. Wait, actually, the maximality is about the chip placement, not the coloring. So perhaps the coloring is maximal in the sense that you can't add any more colored rows or columns? Hmm, maybe not exactly. Let me think again.

Alternatively, the maximal condition on the chip placement translates to the fact that every uncolored row or column cannot be colored without conflicting with existing colors. For example, if there is an uncolored row, then that row must have at least one cell in a column that is colored with both white and black, making it impossible to color the row. Wait, perhaps.

Alternatively, since the configuration is maximal, you can't add any more chips. Therefore, for any uncolored row, if you were to color it white, then all the cells in that row not blocked by columns must be white. But since the configuration is maximal, those cells must already be blocked (i.e., in columns colored black). Similarly, if you tried to color it black, the cells would be blocked by columns colored white. Therefore, for an uncolored row, the columns intersecting it must be colored with both colors? Wait, but a column can only be one color or uncolored. Hmm, perhaps an uncolored row must have all its columns either colored white or black, such that you can't choose a color for the row without conflicting with some column. Wait, if all columns intersecting an uncolored row are colored white, then you could color the row white, so that would not be maximal. Therefore, to prevent that, the uncolored row must have at least one column colored white and at least one column colored black. Therefore, the row cannot be colored either white or black without conflicting with some column. Similarly for uncolored columns.

Therefore, the maximal condition requires that:

- Every uncolored row has at least one column colored white and at least one column colored black in its cells.

- Every uncolored column has at least one row colored white and at least one row colored black in its cells.

Alternatively, if a row is uncolored, then the columns in that row must have both colors, so you can't color the row without conflicting. Similarly for columns.

Therefore, the entire configuration must be such that:

- The colored rows and columns form a grid where their intersections are either filled (if same color) or empty (if different colors).

- Any uncolored row has both colors present in its columns.

- Any uncolored column has both colors present in its rows.

Hmm, this is getting complex. Maybe another approach is better.

Let me think of the grid as being partitioned into four regions:

1. Rows colored white and columns colored white: their intersection is white.

2. Rows colored white and columns colored black: their intersection is empty.

3. Rows colored black and columns colored white: their intersection is empty.

4. Rows colored black and columns colored black: their intersection is black.

Additionally, rows and columns that are uncolored can have cells that are filled with the color of their colored row or column, but if both are uncolored, then those cells can be filled or not? Wait, no. Wait, if a row is uncolored, that means it has no chips, right? Wait, no. Wait, actually, a row is colored only if it has chips. Wait, no, the coloring is determined by the chips. If a row has any chips, they must all be the same color, so the row is considered colored with that color. If a row has no chips, it's uncolored. Similarly for columns.

Therefore, the uncolored rows and columns have no chips. Therefore, the only filled cells are those in colored rows or columns, except where a colored row and colored column of different colors intersect (those are empty). Therefore, the total filled cells are:

For each white row, the number of columns not colored black. Similarly, for each black row, the number of columns not colored white. Wait, perhaps more precisely:

Suppose we have W rows colored white, B rows colored black, w columns colored white, and b columns colored black. The intersections between white rows and white columns are filled white. The intersections between black rows and black columns are filled black. The intersections between white rows and black columns are empty. Similarly, black rows and white columns are empty. The intersections between white rows and uncolored columns are filled white. The intersections between uncolored rows and white columns are filled white. Similarly for black.

Wait, actually, if a column is uncolored, that means it has no chips, so even if a row is colored white, the cell at the intersection of that row and uncolored column can be white (since the column is uncolored, so it doesn't impose a color). Wait, but if the column is uncolored, that means it has no chips. Therefore, if a row is colored white, then all cells in that row must be white (since the row is colored), but if the column is uncolored, that cell can be white (as the column's color is not set). Wait, but the column being uncolored means that it has no chips, so actually, those cells must be empty? Wait, no, that's conflicting.

Hold on, perhaps I need to clarify:

If a row is colored white, that means all the cells in that row that are not blocked by a conflicting column must be white. However, if a column is uncolored, that means it has no chips, so the cells in that column must be empty. Wait, that can't be, because if a row is colored white, then the cells in that row must be white unless blocked by a column. But if a column is uncolored, it has no chips, so the cell at the intersection must be empty. Therefore, if a row is colored white, then all columns that are not colored black must have their cells in that row colored white. But if a column is uncolored, then that cell must be empty, which contradicts the row being colored white (since the row would require that cell to be white). Therefore, there's a mistake here.

Wait, perhaps my initial assumption is wrong. Let me start over.

Let me think of the problem as follows:

Each row can be in one of three states: all white, all black, or empty. Similarly, each column can be all white, all black, or empty. However, the color of a row and column must agree on their intersection if both are non-empty. That is, if row i is white and column j is white, then cell (i,j) is white. If row i is white and column j is black, then cell (i,j) must be empty (since it can't be both). If a row is white and a column is empty, then cell (i,j) can be white (since the column has no constraint). Wait, but if the column is empty, that means it has no chips, so cell (i,j) must be empty. Wait, this is conflicting again.

Ah, here's the confusion. If a column is empty, that means all its cells are empty. Therefore, if a row is colored white, then all cells in that row must be white, but if any column in that row is empty (i.e., the column is entirely empty), then the cell at their intersection must be empty, which contradicts the row being white. Therefore, this is impossible. Therefore, if a row is colored (white or black), then all the columns intersecting it must not be empty. Wait, that can't be, because columns can be empty. Hmm, perhaps the problem is that if a column is empty, then all the cells in that column are empty, so any row going through that column must have an empty cell there, so the row can't be colored. Therefore, in fact, if a column is empty, then every row that passes through it (i.e., all rows) must have an empty cell in that column, so no row can be colored. Wait, that can't be, because if a column is empty, then the rows can still be colored as long as their cells in that column are empty. Wait, but a colored row must have all its non-empty cells of the same color. If a row has some cells filled (in non-empty columns) and some empty (in empty columns), then it's okay, as long as the filled cells are all the same color. Wait, actually, the problem states "all chips in the same row have the same color". So, if a row has any chips, they must all be the same color. The empty cells in the row don't affect this. So, a row can be colored white even if some of its cells are empty (in columns that are empty), as long as all the non-empty cells in the row are white. Similarly for columns.

Therefore, a column can be empty (all its cells empty), and rows can still be colored as long as their cells in non-empty columns are of the same color. So, the key constraints are:

- For any row that is colored (white or black), all its non-empty cells must be of that color.

- For any column that is colored (white or black), all its non-empty cells must be of that color.

- Additionally, if a row is colored white and a column is colored black, their intersection must be empty (since otherwise it would have to be both white and black).

- The arrangement is maximal, so you can't add any more chips. Therefore, every empty cell must be blocked by either its row or column being colored with a different color, or both.

Therefore, to model this, let me consider the following:

Let R be the set of rows colored white, B be the set of rows colored black, C_w be the set of columns colored white, and C_b be the set of columns colored black. The uncolored rows are those not in R or B, and uncolored columns are those not in C_w or C_b.

The cells can be categorized as follows:

1. Cells in R rows and C_w columns: must be white.

2. Cells in R rows and C_b columns: must be empty (since row is white and column is black).

3. Cells in R rows and uncolored columns: must be white (since the row is white, and the column is uncolored, so no conflict).

Wait, but if a column is uncolored, that means it has no chips. Therefore, cells in uncolored columns must be empty. Therefore, cells in R rows and uncolored columns must be empty, which contradicts the row being white. Therefore, this is impossible. Therefore, if a row is colored white, then all columns intersecting it must either be colored white or black? Wait, but if a column is uncolored, then the cell must be empty, but the row requires that cell to be white. Contradiction. Therefore, in fact, if a row is colored, then all columns intersecting it must be colored (either white or black), so that the cells can be appropriately filled or empty. Wait, but columns can be uncolored, but that would require the cells in that column to be empty, conflicting with the row's color.

Therefore, the conclusion is that if a row is colored (white or black), then all columns must be colored (either white or black), otherwise the intersection with an uncolored column would require the cell to be both colored (from the row) and empty (from the column). Therefore, this is impossible. Therefore, in any valid configuration, all columns must be colored if any row is colored, and vice versa? Wait, perhaps not all, but the columns that intersect with colored rows must be colored.

Wait, let me think again. Suppose there is a colored row (say white). Then, for each column in that row, the cell must be either white (if the column is white or uncolored) or empty (if the column is black). However, if the column is uncolored, then the cell must be empty (since the column is uncolored, meaning it has no chips). But the row is white, so that cell should be white. Contradiction. Therefore, if a row is colored, all columns must be colored (either white or black). Similarly, if a column is colored, all rows must be colored. Wait, that can't be, because if you have a colored column, then all rows must be colored to avoid similar contradictions?

Wait, let's suppose a column is colored white. Then, for any row intersecting it, the cell must be white (if the row is uncolored or white) or empty (if the row is black). But if the row is uncolored, then the cell can be white (since the column is white), but the row being uncolored means it has no chips, so the cell must be empty. Contradiction again. Therefore, if a column is colored, all rows must be colored (either white or black). Therefore, in any valid configuration where there is at least one colored row or column, all rows and all columns must be colored. Otherwise, you get a contradiction.

Therefore, the only valid configurations are:

1. All rows and columns are uncolored (i.e., the grid is empty). But this is not maximal, since you can add chips. So this is excluded.

2. All rows and columns are colored, with each row either white or black, each column either white or black, such that for any row and column, if they are colored with different colors, their intersection is empty, and same-colored intersections are filled.

Additionally, the coloring must be such that the arrangement is maximal. Since all rows and columns are colored, the only empty cells are those where a white row and black column intersect, or vice versa. To ensure maximality, we must not be able to add any more chips. Since all rows and columns are colored, any empty cell is at the intersection of a white row and black column (or vice versa). To add a chip there would require violating the color constraints, so that's already blocked. However, could we add a chip to a filled cell? No, since each cell can have at most one chip. Therefore, the configuration is maximal as long as all possible chips that can be placed are already placed. Since the filled cells are exactly those where the row and column colors agree, and the rest are empty, this is indeed maximal because you can't add any more chips.

Therefore, the problem reduces to counting the number of colorings of the rows and columns (each row and column colored either white or black) such that:

- The total number of white chips (cells where row and column are both white) plus the number of black chips (cells where row and column are both black) does not exceed 25 for each color. Wait, but actually, since we have exactly 25 white and 25 black chips available, but the problem says "some of these chips", so we can use any number up to 25 for each, but the configuration must be maximal. However, in our case, since the configuration is determined entirely by the row and column colorings, the number of white and black chips is fixed by the coloring. Therefore, we need to count all such colorings where the number of white cells (agreements between white rows and white columns) is at most 25, and the number of black cells (agreements between black rows and black columns) is at most 25. But since the grid is 5x5, the total number of cells is 25. The white and black cells are disjoint (since a cell can't be both), so the total number of chips is the number of white cells plus black cells, which must be ≤25. But since we can use "some" chips, but the configuration must be maximal. Wait, but in our earlier analysis, the configuration is maximal if and only if all rows and columns are colored (since otherwise you could color an uncolored row or column). Wait, but earlier we saw that if any row or column is colored, all must be colored. Therefore, the only maximal configurations are those where all rows and columns are colored (either white or black), leading to a partition of the grid into four regions: white rows and white columns (white chips), white rows and black columns (empty), black rows and white columns (empty), and black rows and black columns (black chips). The total number of white chips is (number of white rows) × (number of white columns), and similarly black chips are (number of black rows) × (number of black columns). Since the grid is 5x5, the number of white rows plus black rows must be 5, and similarly for columns.

Therefore, let me formalize this:

Let’s denote:

- Let r_w be the number of white rows (0 ≤ r_w ≤5),

- r_b = 5 - r_w (number of black rows),

- c_w be the number of white columns (0 ≤ c_w ≤5),

- c_b = 5 - c_w (number of black columns).

Then, the number of white chips is r_w × c_w,

the number of black chips is r_b × c_b,

and the total chips are r_w c_w + r_b c_b.

However, since the grid is 5x5, the empty cells are r_w c_b + r_b c_w.

The configuration is maximal because all rows and columns are colored, so you can't add any more chips (since any empty cell is at the intersection of a white row and black column or vice versa, so adding a chip there would violate the color constraints).

Therefore, the problem reduces to counting the number of such colorings (i.e., choices of r_w, c_w, and the actual assignments of rows and columns to colors) such that:

1. The number of white chips r_w c_w ≤25,

2. The number of black chips r_b c_b ≤25,

But since the total chips are r_w c_w + r_b c_b, and since the grid has 25 cells, the total chips can be at most 25. However, since we can use "some" chips, but the problem doesn't require using all chips, but the configuration must be maximal. However, in our case, the configuration is maximal regardless of the number of chips, as long as all rows and columns are colored. Wait, but actually, if you have all rows and columns colored, then the configuration is maximal because you can't add any more chips. So even if you have fewer chips (i.e., some rows and columns are colored such that their intersections leave some cells empty but you can't add chips there), it's still maximal. Therefore, all such colorings where all rows and columns are colored (each row and column is either white or black) are valid, provided that the number of white chips (r_w c_w) does not exceed 25 and the number of black chips (r_b c_b) does not exceed 25. But since r_w c_w + r_b c_b ≤25, and each of r_w c_w and r_b c_b are non-negative integers, but actually, since the total is exactly r_w c_w + r_b c_b = total chips, which can be from 0 to 25. However, the problem says "place some of these chips", so as long as the counts are within the available 25 white and 25 black, which they will be automatically since r_w c_w ≤25 and r_b c_b ≤25 because r_w and c_w are at most 5, so their product is at most 25. Wait, actually, 5×5=25, so the maximum for white chips is 25 (if all rows and columns are white), and similarly for black. So the constraints are automatically satisfied. Therefore, all such colorings where all rows and columns are colored (each row and column is either white or black) are valid, and each such coloring corresponds to a unique maximal configuration.

However, we also need to account for the fact that the rows and columns can be assigned colors in different ways. For example, choosing which specific rows are white and which are black, and same for columns.

Therefore, the total number of configurations is the sum over all possible r_w (number of white rows from 0 to 5) and c_w (number of white columns from 0 to 5), of the number of ways to choose the rows and columns:

For each r_w and c_w:

- The number of ways to choose which r_w rows are white (C(5, r_w)),

- The number of ways to choose which c_w columns are white (C(5, c_w)),

- The remaining rows are black (r_b =5 - r_w),

- The remaining columns are black (c_b =5 - c_w).

However, we must ensure that the coloring is consistent. Wait, actually, once you choose the white rows and white columns, the black rows and columns are determined. However, there is a caveat: if both r_w and r_b are positive and c_w and c_b are positive, then the configuration is valid. But actually, there's no conflict here. Each choice of rows and columns colored white or black (with the rest being black) is valid, as long as all rows and columns are colored. Since we are considering all possible colorings where each row is either white or black, and each column is either white or black, then all such colorings are allowed, except for the case where all rows and columns are uncolored (but that's already excluded since we require all rows and columns to be colored for maximality). Wait, actually, in our earlier analysis, the only maximal configurations are those where all rows and columns are colored. Therefore, the total number is the number of ways to color each row and column either white or black, with the only constraint being that you can't have all rows uncolored or all columns uncolored? Wait no, actually, if you color all rows white and all columns white, that's allowed. The only requirement is that all rows and columns are colored (i.e., each row is either white or black, and each column is either white or black). Therefore, the total number is (2^5)*(2^5) = 1024, since each row has 2 choices and each column has 2 choices. However, this counts all possible colorings, including those where, for example, all rows are white and all columns are black, which would result in an empty grid (since all intersections are white rows and black columns). But an empty grid is not maximal, because you could add chips. Wait a second, this is a problem.

Wait, hold on. If all rows are colored white and all columns are colored black, then every cell is at the intersection of a white row and black column, so all cells must be empty. But then the grid is empty. However, an empty grid is not maximal because you can add chips. For example, you could place a white chip in any cell, provided you adjust the row and column colors accordingly. Wait, but in our model, the coloring is fixed. Therefore, in such a configuration where all rows are white and all columns are black, the grid must be empty, but this is not maximal because you could change the color of a column to white, allowing some chips. But in our setup, the coloring is fixed, so perhaps such configurations are considered non-maximal because you can add chips by changing the coloring? Wait, no, the problem states that the placement of chips must be maximal, not the coloring. The coloring is a result of the chip placement.

Therefore, in the case where all rows are white and all columns are black, the chip placement is all empty, which is not maximal. Therefore, such colorings where the resulting chip placement is not maximal must be excluded. Therefore, my previous assumption that all colorings where all rows and columns are colored are maximal is incorrect.

Therefore, I need to refine the analysis.

A configuration (i.e., a coloring of rows and columns) is maximal if you cannot add any more chips without violating the row or column color constraints. However, in the case where all rows are white and all columns are black, the chip placement is empty. You can add a white chip to any cell, provided that the column is also white. But since all columns are black, you can't add a white chip. Similarly, you can't add a black chip because all rows are white. Therefore, actually, in this case, the empty grid with all rows white and columns black is maximal? Wait, because you can't add any chips. Wait, but the problem says "each cell contains at most one chip", so adding a chip to any cell would require that the row and column colors allow it. Since rows are white and columns are black, you can't add any chips. Therefore, that configuration is maximal. But the problem requires that "some of these chips" are placed, but "some" could mean zero? Wait, the problem says "place some of these chips", which might mean at least one. Hmm, the problem statement is a bit ambiguous here. Let me check:

The problem says: "Find the number of ways to place some of these chips in the 25 unit cells..." The term "some" might mean "any number, including zero", but in combinatorics, sometimes "some" means "at least one". However, even if it allows zero, the empty grid is a possible configuration, but we need to see if it's counted.

However, the key point is that even the all-empty grid with all rows white and columns black is considered maximal, because you can't add any chips. Similarly, all-empty with all rows black and columns white is also maximal. However, there's also the all-empty grid where all rows and columns are uncolored, but that's not maximal because you can add chips. Wait, but in our earlier analysis, to be maximal, all rows and columns must be colored. Wait, if all rows and columns are uncolored, then you can add chips, so it's not maximal. But if all rows and columns are colored in a way that blocks all cells, like all rows white and columns black, then it is maximal. So such configurations are allowed.

Therefore, the total number of maximal configurations is indeed all possible colorings of the rows and columns (each row and column colored either white or black), because each such coloring corresponds to a maximal configuration. Because in any such coloring, the empty cells are precisely those where a row and column of opposite colors intersect, and the filled cells are where they agree. Since all rows and columns are colored, you can't add any more chips, so it's maximal. Even if all cells are empty (like all rows white and columns black), that's still maximal. However, the problem says "place some of these chips", which might exclude the all-empty case. 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" could include zero, but the third condition requires maximality. The all-empty grid is maximal because you can't add any chips (since any chip would require its row and column to be colored, but since all rows and columns are colored with conflicting colors, you can't). Wait, actually, in the all-empty grid with all rows white and columns black, you can't add any chips. However, if you have all rows and columns uncolored, then you can add chips, so that's not maximal. Therefore, the all-empty grid is only maximal if all rows and columns are colored in a way that blocks all cells. So the all-empty grid can be achieved in multiple ways: for example, all rows white and columns black, or all rows black and columns white, or any combination where for every cell, the row and column colors conflict.

Wait, actually, the all-empty grid can be achieved in more ways. For instance, if you have some rows white and some black, and columns colored such that every row's color is opposite to the column's color for every cell. But that's only possible if for every row and column, their colors are opposite. Which would require that all rows are the opposite color of all columns. But since rows can be a mix of white and black, this is only possible if all rows are the same color and all columns the opposite. For example, all rows white and all columns black, or vice versa. Therefore, the all-empty grid can only be achieved in two ways: all rows white and columns black, or all rows black and columns white.

Therefore, the total number of maximal configurations is indeed 2^5 * 2^5 = 1024, since each row and column can be independently colored white or black, and each such coloring gives a unique maximal configuration. However, this includes the two all-empty configurations (all rows white/columns black and vice versa). But the problem says "place some of these chips", which might require at least one chip. If so, we need to subtract those two cases. Wait, 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..."

The term "some" might mean "any number, including zero", but in combinatorics, sometimes "some" is used to mean "at least one". However, the problem also says "any additional chip placed on the grid would violate one or more of the previous two conditions". If the grid is empty, then you can place a chip somewhere, provided you set the row and column colors appropriately. Wait, but in the all-empty grid with all rows and columns colored to block all cells (i.e., all rows white and columns black), then you can't add any chips. So that is maximal. However, if you have an all-empty grid with some rows and columns uncolored, then it's not maximal. Therefore, the all-empty grid is only possible in the two colorings where all rows and columns are colored to block all cells. So if the problem requires at least one chip, then those two configurations should be excluded. However, the problem says "some of these chips", which might allow zero. Since the problem includes the two all-empty configurations as valid maximal arrangements, we should include them unless specified otherwise.

But let me see the problem again: it says "place some of these chips". Since they have 25 white and 25 black chips, "some" probably allows any number from 0 to 25 for each color, as long as the conditions are met. Therefore, the total number is 2^5 * 2^5 = 1024. However, this seems too straightforward, and the problem is from a competition, so maybe there's a mistake here.

Wait, but actually, the color assignments of rows and columns are not entirely independent. For example, if you choose some rows to be white and others black, and similarly for columns, but the actual count might have overcounted because of symmetries or something else? Wait, no, each row and column is independently colored, so the total is indeed 2^5 for rows and 2^5 for columns, multiplied together.

Wait, but let me think of a small example. Let's take a 1x1 grid. Then, the number of maximal configurations would be:

Each row (1 row) can be white or black or empty? Wait, no, in the 1x1 case, according to our previous analysis, to be maximal, the row and column must be colored. So the row can be white or black, and the column can be white or black. If they are the same, then the cell is filled with that color. If they are different, the cell is empty. The configurations are:

- Row white, column white: cell is white. This is maximal because you can't add anything (it's already filled).

- Row white, column black: cell empty. Maximal because you can't add a chip (since row is white and column black).

- Row black, column white: cell empty. Similarly maximal.

- Row black, column black: cell black. Maximal.

Additionally, the case where the row and column are uncolored would be the empty cell, but that's not maximal because you can add a chip. Therefore, the maximal configurations are 4, which is 2^1 * 2^1 =4. So that works.

Similarly, for 2x2 grid, it would be 2^2 * 2^2 =16. Let me see if that makes sense. For example, if all rows are white and all columns are white, then all cells are white: that's one configuration. If all rows white and columns black, all cells empty, etc. So yes, 16 seems correct. Therefore, for 5x5, it should be 2^5 * 2^5 = 1024.

However, the problem mentions "indistinguishable" chips. Wait, but the chips are of two colors, but within each color, they are indistinct. However, in our counting above, we considered colorings of rows and columns, which inherently account for the placement of colored chips. Since the chips of the same color are indistinct, the count is correct because we're only counting the color assignments, not the permutations of chips. For example, choosing rows 1-2 as white and columns 1-3 as white gives a certain number of white chips, and since the white chips are indistinct, it's just one configuration.

Therefore, the answer should be 2^5 * 2^5 = 1024. But the problem says "Find the number of ways to place some of these chips...", and since each coloring corresponds to exactly one way (because chips of the same color are indistinct), then 1024 is the answer. However, the problem might have constraints I missed.

Wait, but the problem says "any additional chip placed on the grid would violate one or more of the previous two conditions". In our model, that is satisfied because in any configuration obtained by coloring all rows and columns, you can't add any chips. However, there might be other maximal configurations that are not captured by this model. For example, suppose we have a configuration where some rows and columns are uncolored, but in such a way that you can't add any more chips. Wait, but earlier analysis suggested that if any row or column is uncolored, then you might be able to color it, thereby adding chips. Let me think of an example.

Suppose we have a 2x2 grid. Suppose we color one row white, and one column white, and leave the other row and column uncolored. Then:

- The white row has two cells: one in the white column (filled white), and one in the uncolored column (must be empty because the column is uncolored? Wait, no, the uncolored column has no chips, so that cell must be empty. But the row is white, so that cell should be white. Contradiction. Therefore, such a configuration is invalid. Therefore, indeed, any configuration with uncolored rows or columns is invalid because of the contradictions, so only configurations where all rows and columns are colored are valid. Hence, the total is indeed 2^5 * 2^5 = 1024.

But the problem specifies that there are 25 white and 25 black chips available. However, in our count, some configurations use more than 25 chips of a color? Wait, no. Because in a 5x5 grid, the maximum number of white chips is 5*5=25 (if all rows and columns are white). Similarly for black. So all configurations are within the limits. Therefore, the answer is 1024. But 1024 is 2^10, which is 1024. However, the problem might be expecting a different answer. Wait, maybe I made a mistake in assuming that all colorings are allowed. Let me think again.

Wait, actually, when you choose the colors of the rows and columns, the resulting number of white and black chips must be non-negative, but since they are products of counts, they are automatically non-negative. However, there is another constraint: the white and black chips are separate. For instance, if you have r_w rows white and c_w columns white, then the white chips are r_w*c_w, and the black chips are (5 - r_w)*(5 - c_w). The total chips are r_w c_w + (5 - r_w)(5 - c_w). However, the problem allows using up to 25 of each color, but since the white chips can be at most 25 and black chips at most 25, but since the total is at most 25, both counts are automatically within the limits. For example, if white chips are 25, then black chips are 0, which is okay. If white chips are 24, black could be 1, etc. So all configurations are allowed.

Therefore, the total number is indeed 2^5 * 2^5 = 1024. However, the problem might consider colorings that are equivalent under swapping white and black as distinct, which they are, since the chips are of different colors. So, for example, all rows white and columns black is different from all rows black and columns white. Therefore, the count remains 1024.

But wait, the problem says "indistinguishable white chips and indistinguishable black chips". So the count is correct because we are counting color assignments, and the chips of each color are indistinct. Each coloring corresponds to exactly one arrangement of the chips.

Therefore, I think the answer is 1024, which is 2^10, so 1024. But let me check for a smaller grid again. Let's take 1x1 grid. Then, according to this, the answer would be 4. Let's see:

Possible maximal configurations:

1. White chip: row and column both white.

2. Black chip: row and column both black.

3. Empty with row white and column black.

4. Empty with row black and column white.

So that's 4, which matches 2^1 * 2^1 =4. So that works.

Another test case: 2x2 grid. The total would be 16. Let me see:

Each row can be W or B, each column can be W or B. Each such choice gives a unique configuration. For example:

- All rows W, all columns W: 4 white chips.

- All rows W, all columns B: 0 chips.

- All rows B, all columns W: 0 chips.

- All rows B, all columns B: 4 black chips.

- One row W, one row B; one column W, one column B. Then, the white row and column intersect at 1 cell (white), the white row and black column have 0, the black row and white column have 0, and the black row and column have 1 black. So total 1 white and 1 black chip. There are C(2,1)*C(2,1)=4 such configurations (choosing which row is W and which column is W).

Etc. All total to 16. So that seems correct.

Therefore, for 5x5, the answer should be 2^5 * 2^5 = 1024. However, the problem might have a different answer because of an oversight.

Wait, but the problem says "any additional chip placed on the grid would violate one or more of the previous two conditions". In our model, this is satisfied because in any configuration obtained by coloring all rows and columns, you can't add any chips. However, there might be other configurations that are maximal but don't have all rows and columns colored. Let me think of an example.

Suppose in a 2x2 grid, we have:

- Row 1: white

- Row 2: uncolored

- Column 1: white

- Column 2: uncolored

Then, the filled cells are (1,1) white. The other cells:

- (1,2): row 1 is white, column 2 is uncolored (so column 2 must be empty, so cell (1,2) must be empty, conflicting with row 1 being white. Therefore, this is invalid.

Alternatively, if rows and columns can be uncolored, but then the configuration isn't maximal. For instance, if you have:

- Row 1: white

- Row 2: white

- Column 1: white

- Column 2: uncolored

Then, the cells in column 2 must be empty (since the column is uncolored). But rows 1 and 2 are white, so their cells in column 2 should be white, which contradicts the column being uncolored. Therefore, such configurations are invalid. Therefore, indeed, only configurations with all rows and columns colored are valid, so the count is correct.

Therefore, I think the answer is 1024. But the problem might be expecting something else. Wait, the problem says "indistinguishable white chips and indistinguishable black chips". So when we count the colorings, each coloring corresponds to a unique arrangement, so the count is correct. Therefore, the answer is 1024, which is 2^10, so 1024. But let me see if there's another constraint.

Wait, the problem says "some of these chips", so using zero chips (the two all-empty colorings) might not be allowed. If so, subtract 2, getting 1022. But I'm not sure if "some" requires at least one chip. The problem's third condition requires maximality. The all-empty configurations with all rows and columns colored are maximal, so they should be included unless the problem specifies non-empty. Since the problem says "some", which can include zero, I think they should be included. Therefore, 1024 is the answer.

However, the problem might have intended that the grid must be non-empty, so perhaps subtract 2. But since the problem didn't specify, I think the correct answer is 1024. Wait, but let me think again.

Wait, in the problem statement, it says "place some of these chips", which could mean that you have to place at least one chip. In that case, we need to subtract the two all-empty configurations (all rows white and columns black, and vice versa). So the total would be 1024 - 2 = 1022. Hmm, but which is it?

Looking back at the problem statement:

"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" here might mean "one or more", so the all-empty configurations are excluded. Because "placing some chips" could imply at least one. In that case, the answer would be 1024 - 2 = 1022. However, in combinatorics, "some" can sometimes include zero. To check, let's see if the all-empty grid is considered a valid "placement" here. The problem says "place some of these chips", so if "some" allows zero, then it's included; otherwise not.

Alternatively, maybe the problem considers that "placing some chips" requires that you use at least one chip of each color? But that's not indicated.

Alternatively, perhaps my entire approach is wrong. Maybe the problem requires that in every row and column that has chips, they must be all the same color, but rows and columns can be partially filled. Wait, but 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. So a row can be partially filled, but all its chips must be the same color. Similarly for columns. However, the maximal condition complicates things.

Wait, perhaps my initial approach was incorrect in assuming that all rows and columns must be colored. Let me try a different approach.

Suppose that in a maximal configuration, every row that is not entirely filled must have its color blocked by the columns, and vice versa.

Alternatively, think of the grid as being partitioned into monochromatic rows and columns, such that adding any chip would require changing a row or column's color, which is not allowed.

Alternatively, the maximal condition implies that every empty cell is "blocked" by its row or column. For a cell to be empty, either its row is already filled with the opposite color, or its column is, or both.

Wait, let me think of the problem as a bipartite graph between rows and columns, where an edge represents a cell. Coloring a row or column imposes a color on the edges (cells) incident to it, with conflicts forbidden.

Alternatively, here's another way: the configuration is determined by choosing for each row a color (W, B, or empty), and similarly for each column, with the constraints:

- If a row is colored W, then all its non-empty cells must be W, and similarly for B.

- If a column is colored W, all its non-empty cells must be W.

- The coloring must be maximal: cannot add any more chips.

However, this seems to loop back to the earlier analysis.

Alternatively, perhaps the problem is equivalent to choosing a subgrid that is a rectangle of white chips and a rectangle of black chips, such that they don't overlap and cover the grid maximally. Wait, maybe not.

Alternatively, think of the grid as having some rows colored white and some black, and some columns colored white and some black. The intersection of a white row and white column is white, black row and black column is black, and intersections of different colors are empty. The rest of the cells (in uncolored rows or columns) can be filled with the row or column's color. Wait, but earlier analysis said that uncolored rows/columns can't exist in a maximal configuration.

Alternatively, the key idea is that the maximal configurations are exactly those where the grid is partitioned into four rectangles: white rows intersect white columns (white chips), black rows intersect black columns (black chips), and the other two rectangles are empty. The counts of white and black chips are then (number of white rows)*(number of white columns) and (number of black rows)*(number of black columns). The total is the sum, which must be ≤25 for each color. Since the grid is 5x5, the numbers work out.

Therefore, the number of such configurations is the number of ways to choose:

- A subset of rows to be white (the rest are black),

- A subset of columns to be white (the rest are black).

Each such choice gives a unique configuration. The number of subsets for rows is 2^5, and similarly for columns, so total 2^5 * 2^5 = 1024. This includes the two all-empty cases (all rows white and columns black, or vice versa). If the problem requires at least one chip, subtract 2.

However, the problem says "some of these chips", which could include zero. Since the problem does not specify that at least one chip must be placed, I think we should include all cases. Therefore, the answer is 1024.

But the problem is from a competition, perhaps the answer is different. Wait, maybe I made a mistake in assuming that the rows and columns can be independently colored. For instance, if you choose some rows to be white and others black, and independently do the same for columns, but the actual color assignments must be consistent in some way?

Wait, no, the colors are independent. Each row can be chosen as white or black independently, and same for columns. The only constraint is that the intersections of differently colored rows and columns are empty, but that is automatically handled by the coloring. Therefore, each such coloring is a valid configuration, and they are all maximal.

Therefore, I think the correct answer is 1024. However, since the problem mentions "indistinguishable" chips, perhaps we need to account for overcounting? Wait, no, because the chips of the same color are indistinct, but the configurations are distinguished by their color assignments. For example, choosing rows 1-2 as white and columns 1-3 as white is a different configuration from rows 1-3 white and columns 1-2 white, even though they might have the same number of chips. Since the problem counts the number of ways to place the chips, considering different arrangements as different even if they have the same number of chips in different positions, the count is correct.

Therefore, I think the answer is 1024, so 2^10 = 1024. Therefore, the final answer is \boxed{1024}.

Wait, but hold on. Wait a second. Let me think again. The problem says "indistinguishable white chips and indistinguishable black chips". So when we count the number of ways, it's the number of distinct colorings of the grid where each cell is either white, black, or empty, subject to the constraints. However, in our model, each choice of row and column colors gives a unique grid coloring. Since the white and black chips are indistinct, but their positions matter, the count is correct as 1024.

Alternatively, perhaps the problem requires that all chips of the same color are placed in contiguous blocks or something, but the problem doesn't say that. The only constraints are on the rows and columns having uniform color.

Therefore, after careful consideration, I believe the answer is 1024.

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

We are given a $5 \times 5$ grid and need to place some of the 25 indistinguishable white chips and 25 indistinguishable black chips such that each cell contains at most one chip, all chips in the same row and column have the same color, and the placement is maximal (no additional chip can be placed without violating the previous conditions).

Key steps and reasoning:

1. **Row and Column Coloring**: Each row and each column must be colored either white, black, or left uncolored. However, for the placement to be maximal, every row and column must be colored (either white or black). This is because an uncolored row or column would allow adding chips, violating maximality.

2. **Maximal Configuration**: A configuration is maximal if every empty cell is at the intersection of a white row and black column or vice versa. This ensures no additional chip can be placed without violating the color constraints.

3. **Counting Configurations**: Each row can be independently colored white or black, and each column can also be independently colored white or black. There are $2^5$ ways to color the rows and $2^5$ ways to color the columns. Thus, the total number of configurations is $2^5 \times 2^5 = 1024$.

4. **Inclusion of All Configurations**: This count includes all possible colorings, including those where the grid is entirely empty (but maximally so, such as all rows white and columns black or vice versa). The problem allows "some" chips, which can include zero, so these configurations are valid.

Final Answer:
\boxed{1024}