If 3 squares are chosen at random from the 64 | vedclass

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(C) The total number of ways to choose $3$ squares from $64$ is $\binom{64}{3} = \frac{64 	imes 63 	imes 62}{3 	imes 2 	imes 1} = 41664$. On a che

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$\frac{7}{744}$

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(C) The total number of ways to choose $3$ squares from $64$ is $\binom{64}{3} = \frac{64 	imes 63 	imes 62}{3 	imes 2 	imes 1} = 41664$. On a chess board,the diagonals have lengths $1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1$ in both directions. For a diagonal of length $k$,the number of ways to choose $3$ squares is $\binom{k}{3}$. We only consider diagonals with length $k \ge 3$. The lengths are $3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3$. There are two sets of these diagonals (one for each direction). Sum of $\binom{k}{3}$ for one direction: $\binom{3}{3} + \binom{4}{3} + \binom{5}{3} + \binom{6}{3} + \binom{7}{3} + \binom{8}{3} + \binom{7}{3} + \binom{6}{3} + \binom{5}{3} + \binom{4}{3} + \binom{3}{3} = 1 + 4 + 10 + 20 + 35 + 56 + 35 + 20 + 10 + 4 + 1 = 196$. Since there are two directions,the total number of favorable outcomes is $196 	imes 2 = 392$. The probability is $\frac{392}{41664} = \frac{7}{744}$.

If $3$ squares are chosen at random from the $64$ squares of a chess board,then the probability that all of them lie along the same diagonal line is

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