Question

n different toys are to be distributed among n children. Find the number of ways in which these toys can be distributed so that exactly one child gets no toy

Answer 1

$1$ child gets $0$ toys, $(n-1)$ children get $1$ toy each, any of the $(n-1)$ children can get $2^{nd}$ toy.

$1$ child can be left out a toy in $n$ ways.The 'extra' toy can be chosen in $n$ ways and so can be distributed among the remaining $(n-1)$ children in $n(n-1)$ ways.The remaining $(n-1)$ toys can be distributed among the remaining $(n-1)$ children (one toy each ) in $(n-1)!$ ways.

However, if the child getting $2$ toys gets toy $A$ as the 'extra' toy and toy $B$ as the 'ordinary' toy, this is the same as if this child gets toy $B$ as the extra toy and toy $A$ as the ordinary toy. So we have counted $2x$ as many combinations as we need.

So the number of distributions is $=n\times n(n-1)\times \frac{(n-1)!}{2}$

$=n!\times \frac{n(n-1)}{2}$

$=n!\frac{n!}{2!(n-2)!}$

$=n!\times C(n,2)$