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?

Each of the 64 squares of the chessboard has 7 other squares in its same row, and 7 other squares in its same column.

So, for each of the 64 rows in which we can place the first piece, we have 14 choices for the square for the second piece.

But the pieces are indistinguishable, so we have to divide that number by two, because each combination has a "twin" one, where you flip the two pieces, obtaining the exact same combination.

So, the answer is

true !

step-by-step explanation: