Question

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 each in a box, could be placed such that a ball does not go to a box of its own colour is _______.

Answer 1

We know that, if $n$ different things are arranged in a row,
then the number of ways in which they can rearranged so that none of them occupies its original place is $n!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{{(-1)}^n}{n!})$
Now assume that each boll is placed in the box of its own choice and apply the above result
Hence the required number of different ways is
$4!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})=\frac{4!}{2!}-\frac{4!}{3!}+1$
$=12-4+1=9$