In how many ways can $n$ toys be distributed to $n$ children so that exactly one child does not receive a toy? $n$ toys are to be distributed amongst $n$ children. The total number of ways in which these toys can be distributed so that exactly one child gets no toy is equal to?
I tried it with the concept that if one child gets no toy, then we are left with $n$ toys to be distributed amongst $n-1$ children. But this does not help and the answer is not coming. 
Please somebody,can you give the detailed explanation how to solve this sum. Thanks
 A: The question is stated as follows:

If $n$ $\color{red}{different}$ toys have to be distributed among $n$ children, then the total number of ways in which these toys can be distributed so that exactly one child gets no toy is equal to?

Source: link
Step 1: There are $n$ ways to choose the unlucky child.
Step 2: There are ${n\choose 2}$ ways to choose $2$ toys. Note: only 1 child can get maximum $2$ toys, otherwise more than $1$ child will get no toy.
Step 3: There are $n-1$ ways to choose the luckiest child, who will get the $2$ toys.
Step 4: There are $(n-2)!$ ways to distribute the rest $n-2$ toys to the rest $n-2$ children.
Hence:
$$n\cdot {n\choose 2}\cdot (n-1)\cdot (n-2)!=n!{n\choose 2}.$$
A: Judging by the answer you provided in the comments, the toys must be distinct.  Therefore, we wish to count the number of ways $n$ different toys can be distributed to $n$ children if exactly one child does not receive a toy.
Strategy: 


*

*Choose which of the $n$ children receives no toy.

*If the $n$ toys are distributed to the remaining $n - 1$ children, exactly one of those $n - 1$ children will receive two toys.  Choose that child.

*Choose which two of the $n$ toys that child receives.

*Give each of the remaining $n - 2$ children one of the remaining $n - 2$ toys.

A: $1$ child gets $0$ toys, $(n-1)$ children get $1$ toy each, any of the $(n-1)$ children can get a $2$nd toy.
$1$ child can be left without 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 $2\times$ as many combinations as we need.
So the number of distributions is
$$  \frac{n  n(n-1)(n-1)! }{ 2}
= n!  \frac{n(n-1)}{2}
= n! \frac{ n!}{2!(n-2)!}
= n!  C(n,2). $$
