Question

There are four balls of different colors and four boxes of colors 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 color is .....

Answer 1

We know that number of derangements of n objects is
$n![1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...+\frac{1}{n!}]$
Therefore, number of ways of putting all the 4 balls into boxes of different colour is
$4![1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}]=4!(\frac{1}{2}-\frac{1}{6}+\frac{1}{24})$
$=24\left(\frac{12-4+1}{24}\right)$
$=9$