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 in each box, could be placed such that a ball doesn't go to box of its own colour is

• A. 8
• B. 7
• C. 9
• D. None of these

Answer 1

Answer: (C)

9

Answer 2

Number of derangement are = 4 !

$\left\{ \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right\} = 12 - 4 + 1 = 9$.

(Since number of derangements in such a problem is given by $n!\left\{ 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}...... + ( - 1)^{n}\frac{1}{n!} \right\}$.