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

• A. $7$
• B. $8$
• C. $9$
• D. $20$

Answer 1

Answer: (C)

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 ball is placed in the box of its own color and apply the above result.
Hence, the required number of different ways is.
$4!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})=\frac{4!}{2!}-\frac{4!}{3!}+1$
$=12-4+1=9$