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If three one by one squares are drawn from the chessboard then the probability that they form the letter "L" is?

I was thinking that if we select a 4 sided square is selected it will contain an "L".Am i going on right track?Please help!

  • Very clever! You are on the right track (a track I would never have guessed myself). Wonderful insight. Just think: how many $2 \times 2$ squares are on the board? And how many L shapes per square? Finally, how many ways can you choose $3$ objects from $64$? – Colm Bhandal Dec 10 '15 at 16:56
  • Also an important property is that an L shape uniquely identifies the square to which it belongs, so you're not allowed have $2$ distinct squares containing the same L shape. This allows you to proceed as above. – Colm Bhandal Dec 10 '15 at 16:59

2 Answers2

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Hint: We assume that the chessboard is the conventional $8\times 8$, and that we have to see an L when we are sitting in the usual position, so an "upside down" L doesn't count.

There are $\binom{64}{3}$ equally likely ways to choose $3$ squares. Now we count the number of choices that give an L.

The top square of the L can be in any one of the top $7$ rows. How many ways are there to place an L with top square in one of these rows? Your "filling out the L to a square" is a good way to visualize.

André Nicolas
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For a Chessboard you have 64 distinct squares, from which 3 squares can be selected in 64C3 ways, which is the total number of possible outcomes. Now to get the required pattern in a fixed orientation, the top square which makes the "L" can be chosen in 49 ways since the 8th row and the 8th column cannot house the top square. That makes exactly 15 squares(8+8-1)which are out of our choice list. So the first square can be chosen from any one of the 49(64-15) squares. Once the first square is chosen there is only one option to choose the remaining two squares to complete the pattern. That makes the probability 49/(64C3).

Aneesh Ghosh
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