Three different squares are chosen randomly on a chessboard.
What is the probability that they lie in the same diagonal?
I got the same answer as Charlie, I used nothing more than a calculator for factorial. Perhaps this is a simpler way of looking at it, as it requires no conditional probabilities or programs.
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392/41664 = .009408602 (or a little less than 1% chance)
There are [ (64 x 63 x 62) / (3 x 2 x 1) ] = 32 x 21 x 62 =
41664 different combinations of choosing 3 squares at a time.
Along a NW - SE direction (on the chessboard), there are 15 diagonals, 4 of which have only 1 or 2 squares, leaving 11 diagonals that could potentially contain the 3 random squares.
Those 11 diagonals contain:
3, 4, 5, 6, 7, 8, 7, 6, 5, 4, and 3 squares to contain the 3 random squares.
Those diagonals can each contain that many "choose 3" at a time possible combinations of the three random squares, so respectively:
3!/(3!0!), 4!/(3!1!), 5!/(3!/2!), 6!/(3!3!), 7!/(3!4!), 8!/(3!5!), and then back down again... which gives us:
1, 4, 10, 20, 35, 56, 35, 20, 10, 4, and 1
This totals (1 + 4 + 10 + 20 + 35 + 56 + 35 + 20 + 10 + 4 + 1) =
196.
We must also allow for the diagonals going from NE - SW, and they will also have
196 possible combinations.
So, there are 196 x 2 = 392 possible locations.
392 possible locations that work / 41664 different combinations of placing 3 squares at a time =
392/41664 = .009408602 (or a little less than 1% chance)
solution, simplification?
