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(D) chess board has $32$ black and $32$ white squares.Total ways to choose one white and one black square $= {}^{32}C_{1} 	imes {}^{32}C_{1} = 32 	i

Vedclass · Accepted Answer

$768$

Vedclass · Answer

(D) chess board has $32$ black and $32$ white squares.Total ways to choose one white and one black square $= {}^{32}C_{1} 	imes {}^{32}C_{1} = 32 	imes 32 = 1024$.There are $8$ rows,and each row contains $4$ white squares and $4$ black squares.Ways to choose a white and a black square in the same row $= 8 	imes ({}^{4}C_{1} 	imes {}^{4}C_{1}) = 8 	imes 16 = 128$.Similarly,there are $8$ columns,and each column contains $4$ white squares and $4$ black squares.Ways to choose a white and a black square in the same column $= 8 	imes ({}^{4}C_{1} 	imes {}^{4}C_{1}) = 8 	imes 16 = 128$.Total ways where squares lie in the same row or column $= 128 + 128 = 256$.Therefore,the required number of ways $= 1024 - 256 = 768$.

In how many ways is it possible to choose a white square and a black square on a chess board so that the squares do not lie in the same row or column?

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