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The total number of possible ways to choose 3 squares from a $8\times 8$ chessboard: $\displaystyle N = { 64 \choose 3} = \frac {64\times 63 \times 62} {1\times 2\times 3} = 41664$

There are two sets of 15 diagonals (positive and negative slopes or incline to the right and left respectively) that cut through the following numbers of squares: $\{ 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1\}$

We note that if the diagonal cuts through $n$ squares, the number of ways to choose three squares on the diagonal is given by $\displaystyle {n \choose 3}$ . Therefore, the total number of possible ways to choose 3 squares on all 30 diagonals is given by:

$\begin{aligned} M & = 2 \left[{1 \choose 3} + {2 \choose 3} + {3 \choose 3} + {4 \choose 3} + {5 \choose 3} + {6 \choose 3} + {7 \choose 3} + {8 \choose 3} + {7 \choose 3} +{6 \choose 3} + {5 \choose 3} + {4 \choose 3}+ {3 \choose 3} + {2 \choose 3} + {1 \choose 3} \right] \\ & = 2 (0+0+1+4+10+20+35+56+35+20+10+4+1+0+0) \\ & = 2 (196) = 392 \end{aligned}$

Therefore the probability that the 3 squares chosen line on a diagonal: $\dfrac MN = \dfrac {392}{41664} = \boxed {\dfrac {7}{744} }$

Total Number of ways to choose 3 squares = $64 \times 63 \times 62$

As there are 64 ways to choose the first square, there are 63 ways to choose the second square and so on.

Diagonal with size 8 appears 2 times and the diagonal with size {3, 4, ..., 7} appears 4 times.

$\therefore$ Number of times we get 3 squares that lie on a diagonal out of total outcomes is

$2 \times {}^8\!P_3 + 4 \times \sum_{i = 3}^7 {}^i\!P_3\\ = 2 \times 336 + 4 \times (6 + 24 + 60 + 120 + 210) = 2 \times 336 + 4 \times 420 = 2352$

$\therefore$ The probability to get the three squares on a diagonal $= \frac{2352}{64 \times 63 \times 62} = \frac{7}{744}$