Question

There are six balls of different colours and six boxes of colours same as those of the balls. The number of ways in which the balls, one in each box, could be placed in the boxes such that atmost two balls are in their corresponding colour boxes is equal to

Answer 1

Given, six balls of different colours and six boxes of similar colour.
Condition : atmost two balls are in their corresponding colour boxes.

Case-I : No ball is in their corresponding colour box.
Number of ways
$=6!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!})$
$=265$

Case-II : One ball is in the corresponding colour box.
Number of ways
$={}^6{C}_1\cdot 5!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!})$
$=6\times 44=264$

Case-III : Two balls are in their corresponding colour boxes,
Number of ways.
$={}^6{C}_2\cdot 4!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})$
$=15\times 9=135$

$\therefore$ Total number of ways $=265+264+135=664$