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If $5$ squares are chosen at random from a chess board, what is the probability that they lie on a diagonal line?

Matthew Conroy
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juantheron
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2 Answers2

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HINT:

  1. There are $64$ squares altogether; how many ways are there to choose $5$ of them?

  2. There are $15$ diagonals in each direction, but only $7$ of them are long enough to contain $5$ squares. Specifically, in each direction there are two diagonals of length $5$, two of length $6$, two of length $7$, and one of length $8$. How many $5$-element subsets are there of each of these diagonals?

The total in (2) is the number of sets of $5$ squares that lie on a diagonal. The total in (1) is the total number of sets of $5$ squares. Combine these two numbers to get the desired probability.

Brian M. Scott
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  • Thanks Brain M. Scoot Got it.We can select $5$ square out of $64$ is $\displaystyle \binom{64}{5}$ now for favourable no. of cases $ = 2.\left{\binom{5}{5}+\binom{6}{5}+\binom{7}{5}\right}+\binom{8}{5}$ This is for anong one diagonal $(AC)$ same things for another diagonal $(BD)$ So tatal is $ = 112+112 = 224$ So probability is $\displaystyle = \frac{224}{\binom{64}{5}}$ – juantheron Jan 29 '13 at 19:26
  • @juantheron: You’re welcome; looks good. – Brian M. Scott Jan 29 '13 at 19:28
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There is $16$ possible arrangements of $5$ squares so that they are on the diagonal lines and $64*63*62*61*60$ possible ways to put squares on the chessboard so the probability is $16\over(64*63*62*61*60)$

A.P.
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  • While I suppose the denominator could be correct (taking the order in which squares are chosen as making a difference), there's no way the numerator 16 could be compatible with this way of counting. Even one set of five squares in a diagonal would give rise to a count of at least 5! arrangements of those squares (according to the order chosen). – hardmath Jan 29 '13 at 19:18
  • Well, it depends on whether you see $5$ chosen squares which are on the same position as one arrangement or $5!$ arrangements, i choose it to be one because they constitute the same location on the chessboard. – A.P. Jan 29 '13 at 19:27
  • Then your enumeration of all subsets of 5 squares on the board (denominator) is also wrong. – hardmath Jan 29 '13 at 21:06