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

Question

I am a student in middle school and I was wondering if anyone could help me with the following problem:

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

I think there would be 64 options for the first piece since it can go anywhere on the chessboard, but I'm not sure about the second piece.

Answer 1

First of all, let's choose one of the two possibilities: either the same row OR the same column. They can't be both because then the two pieces would be the same. There are $16$ total rows and columns.

Next, given a single row/column, we choose two squares out of that specific row/column.

There are 8 squares in that row/column, and we want two of them for a total of $\binom{8}{2}$ or $28$.

Thus, the final answer is $$\text{number of ways to choose a row/column}\times\text{number of ways to choose the squares} $$ $$= 28\cdot 16 = 448$$

Does this answer your question?

Answer 2

The first piece can go in any of the $64$ squares. The second piece can then be in any of $14$ positions, since there are $7$ unoccupied squares in the row of the first piece, as well as $7$ unoccupied squares in the column of the first piece. This would seem to give us $64\cdot 14$ choices for the placement of the two pieces. However, order doesn't matter (we said the pieces are indistinguishable), so the actual number of choices is $(64\cdot 14)/2$, which is $\boxed{448}$.