In how many ways can we choose two black squares on a chess board

Method 1:

Total number of black squares on the board = 32
Number of ways of selecting the first black square = \(32C_1\) = 32
The selected black square will be common in the row and column from which we cannot select the second black square,
Therefore, the number of black squares unavailable for selecting the second = 4 in the row which cannot be used + 4 in the column which cannot be used - 1 common = 7
Therefore, number of black squares available for selecting the second = 32-7 = 25
Number of ways of selecting the second black square = \(25C_1\) = 25

Therefore, total number of ways of selecting the 2 black squares = 32*25/(No. of repeating colours!) = 800/2! = 400 (we divide because the two squares are both black and hence the order of selection doesn't matter here)

Answer (B)

Method 2:

Number of ways of selecting 2 black squares from the same row = \(4C_2*8\) = 48
Number of ways of selecting 2 black squares from the same column = \(4C_2*8\) = 48
Total number of incorrect ways of selecting 2 black squares (ie; such that they either lie in the same row or the same column) = 48 + 48 = 96
Total number of ways of selecting 2 black squares = \(32C_2\) = \(\frac{32*31}{2*1}\) = 496

Therefore, total number of ways of selecting the 2 black squares such that they don't lie in the same row or column = 496 - 96 = 400

Answer (B)

Hope this helps.