Question

There are $4$ balls of different colours and $4$ boxes of same colours as 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 box of its own colour is

• A. $8$
• B. $9$
• C. $7$
• D. $10$

Answer 1

The correct option is B $9$

Let $B_1,B_2,B_3,B_4$ be balls and

$A_1,A_2,A_3,A_4$ be the respective boxes of respective colour.

Let ball $B_1$ does not go in box $A_1$ which is of some colour of that of the ball.

i.e, $B_1$ can be placed in any of three boxes $A_1,A_2,A_3$

If $B_1$ is placed in box $A_2$, then the other 3 balls can be placed in the following 3 ways.

$\begin{array}{cccc}{A}_1& B_2& B_3& B_4\\ A_3& B_4& B_4& B_2\\ A_3& B_3& B_2& B_3\end{array}$

Similarly if $B_1$ is placed in box $A_3$ or in $A_4$ we get 3 ways in each case.

$\therefore$ The required number of ways $=3\times 3=9$