There are 4 balls of different colours and 4 boxes of the same colours as those of the balls respectively. Find the number of ways to put one ball in each box so that only two balls are in boxes with respect to their colour.

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Solution

The correct option is C 6
The required number of ways
$= \frac{4!}{2!} [1 - \frac{1}{1!} + \frac{1}{2!}] = 4 \times 3 [1 - 1 + \frac{1}{2}] = 6$

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There are 4 balls of different colours and 4 boxes of the same colours as those of the balls respectively. Find the number of ways to put one ball in each box so that only two balls are in boxes with respect to their colour.

The correct option is C 6The required number of ways =4!2![1−11!+12!]=4×3[1−1+12]=6

The correct option is C 6
The required number of ways
$= \frac{4!}{2!} [1 - \frac{1}{1!} + \frac{1}{2!}] = 4 \times 3 [1 - 1 + \frac{1}{2}] = 6$