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Q. There are four balls of different colours and four boxes of colours same as those of the balls. The number of ways in which the balls, one in each box, could be placed such that a ball does not go to box of its own colour, is

Permutations and Combinations

Solution:

The number of ways in which four different balls can be placed in four different boxes
$ = \,{}^{4}C_{1} +\,{}^{3}C_{1} +\,{}^{2}C_{1} +\,{}^{1}C_{1} = 4 + 3 + 2 + 1 = 10$
$\therefore $ Required number of ways $= 10 - 1 = 9$
(since, only one way in which the same ball have a same box).