How many ways can I put down two indistinguishable pieces on an ordinary 8x8 chessboard, if the pieces must either be in the same row or be in the same column?

Umeko1.0 Apr 2, 2024

#1**0 **

There are two cases to consider for placing the two indistinguishable pieces on the chessboard:

Same Row: The pieces can be placed on any of the 8 rows. Once the row is chosen, either piece can be placed on any of the 8 squares in that row.

So, for the same row scenario, there are 8 ways to pick a row and 8 ways to arrange the pieces within that row, resulting in 8 * 8 = 64 arrangements.

Same Column (excluding diagonal): The pieces can be placed on any of the 8 columns (excluding the possibility of them being on the same diagonal, which we'll address later).

Similar to the same row case, once a column is chosen, either piece can be placed on any of the 8 squares in that column.

However, this counts the diagonal placement twice (once for each way of choosing the row or column first).

So, for the same column scenario (excluding diagonal), there are 7 ways to pick a column and 7 ways to arrange the pieces within that column, resulting in 7 * 7 = 49 arrangements.

Correcting for Diagonal Placement:

We've double-counted the arrangements where the pieces are on the same diagonal.

How many are there? Imagine choosing the diagonal first. There are only 4 choices (one for each corner).

Then, you can place the pieces on two squares out of the 8 on that diagonal, giving 8 choices.

However, since we've already counted this scenario twice (once for each way of choosing the row or column first), we need to subtract 1.

Therefore, the number of diagonal placements counted twice is 4 (diagonals) * (8C2) (combinations to choose 2 squares out of 8) - 1 (double counting) = 14.

Total Arrangements:

Adding the arrangements for same row, same column (excluding diagonal), and subtracting the double-counted diagonals, we get the total number of possible placements:

Total = Same Row + Same Column (excluding diagonal) - Double Counted Diagonals Total = 64 + 49 - 14 Total = 99

There are 99 ways to place two indistinguishable pieces on an 8x8 chessboard if they must be in the same row or the same column.

bader Apr 2, 2024