Two squares of `1xx1` are chosen at random on a chestboard. What is the probability that they have a side in common ?

Options

A. `1//18`
B. `64//4032`
C. `63//64`
D. `1//9`

Answer

A

Explanation

`(a)` Total number of ways of selecting two squares
`"^(64)C_(2)=(1)/(2)(64xx63)=2016`
If the first square happens to be any of the four corner ones, the second square can be chosen in `2` ways.
If the first square happens to be any of the 24` squares on the side of the chess board, the second square can be chosen in `3` ways.
If the first square happens to be any of the `36` remaining squares, the second square can be chosen in `4` ways.
Hence the required number of combinations
`=(1)/(2)((4xx2)+(24xx3)+(36xx4))=112`
Therefore, the required proability `=1//18`.