Question

There are $5$ balls of different colours and $5$ boxes of colours same as those of the balls. The number of ways in which the balls, one in each box, could be placed such that $2$ ball goes to the box of its own colour.

Answer 1

Let $B_1,B_2,B_3,B_4,B_5$ be balls and $A_1,A_2,A_3,A_4,A_5$ be the respective boxes of respective colour.

Let Ball $B_1$ does nit in box $A_1$, which is of same colour of that of the ball.

i.e. $B_1$ can be placed in any of the three boxes / four boxes. $A_1,A_2,A_3$ & $A_4$.

If $B_1$ is placed in box $A_2$, then outer 4 balls can be placed in fall. 4 ways.

$A_1$	$B_2$	$B_3$	$B_4$	$B_5$
$A_3$	$B_4$	$B_4$	$B_2$	$B_3$
$A_4$	$B_3$	$B_2$	$B_3$	$B_2$
$A_2$	$B_2$	$B_3$	$B_4$	$B_2$

$\therefore$ No. of required ways = $13\times 5=65$ ways