Permutations & Combinations on CHESSBOARD/ GRID.....

you may get to the answer by eliminating options itself. only 400 seems close to the answer (it has to be greater than 32 - because that is the number of black squares on the board. right?)

anyhow, the solution:

number of black squares on the board = 32 (4 in each of the 8 rows - 4 in each of the 8 columns)

ways of selecting 1 black square out of 32 = 32
that makes us exclude 7 black squares for our next selection (3 each from the row and the column we picked our first black square and that selected black square itself)
hence, ways of selecting the second black square = 32-7=25

so total ways = 32x25=800

now because the 2 black squares are identical, you have counted them twice in your calculation. you have counted cases for the same pair where a black box was considered first selection and was considered the second selection separately.

we want selection and not permutation, so divide by 2. you get 400.

did that help?

Total 32 black squares
4 black squares in each row and each column


1st square can be chosen in 32 ways.
2nd square can be chosen in 25 ways = 32 - (1 chosen + 3 in same row + 3 in same column)
total = 32x25 = 800.
Am I missing something ?

Did you get the last para of the previous post?
Basically, you need to divide by 2 because you have counted each pair twice.

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