Question

There are four balls of different colors and four boxes of colors same a 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 a box of its own color is?

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

Answer 1

The correct option is C

$9$

Explanation of the correct option

We know that, for $\mathrm{n}$ different things arranged in a row, number of ways of arranging them such that none of them takes its original place, is given by:

$\mathrm{n}!\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...................\frac{{\left(-1\right)}^{\mathrm{n}}}{\mathrm{n}!}\right)$

Therefore, the required number of ways is

$\begin{array}{rcl}& =& 4!\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}\right)\\ & =& \frac{4!}{2!}-\frac{4!}{3!}+1\\ & =& 12-4+1\\ & =& 9\end{array}$

Hence option $\left(\mathrm{C}\right)$, $9$ is the correct answer.