Question

Suppose there are 7 boxes of different colours each containing a ball of same colour as of box. If the number of possible ways of arranging these balls in the boxes (one in each) so that exactly three balls are in the box of their own colour is m. Then $\frac{m}{5}$ equals to

Answer 1

63

Answer 2

The problem requires finding the number of ways to arrange 7 distinct balls into 7 distinct boxes such that exactly 3 balls are in their corresponding colored boxes.

This involves two steps:

1. Choosing the balls in their correct boxes: The number of ways to choose which 3 out of 7 balls go into their own colored boxes is given by the combination formula: $\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$
2. Deranging the remaining balls: The remaining $7 - 3 = 4$ balls must be placed in the remaining 4 boxes such that none are in their own colored box. This is a derangement problem. The number of derangements of 4 items, $D_4$, is calculated as: $D_4 = 4! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right) = 24 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right) = 9$

The total number of ways, $m$, is the product of these two values: $m = \binom{7}{3} \times D_4 = 35 \times 9 = 315$ Finally, we need to calculate $\frac{m}{5}$: $\frac{m}{5} = \frac{315}{5} = 63$