Prove that ${n\choose 2}2^{n-2}=\sum\limits_{k=2}^{n}{n\choose k}{k\choose 2}$  for all $n\geq 2$

Use a combinatorial argument to show that 

                                                      ${n\choose 2}2^{n-2}=\sum\limits_{k=2}^{n}{n\choose k}{k\choose 2}$ for all $n\geq 2.$

Answer

Assume that there if going to be a party and you have $n$ friends, and $2$ out of your $n$ friends will go and bring pizza. The remaining $n-2$ may or may not go to the party. The total number of ways that your friends may go to the party and bring pizza is 
\[{n \choose2}2^{n-2}.\]
Indeed there are ${n \choose2}$ ways to choose the two friends who would bring pizza, and the remaining $(n-2)$ each have 2 options i.e. they may or may not go to the party which gives $2^{n-2}$ possibilities.

We can count the total number of ways that your friends may go to the party and bring pizza in a different way. A total number k, $k\geq 2$, may attend the party and two of them will bring pizza. The number of possibilities are
\[{n \choose k}{k \choose2}.\]
Summing over $k$ we get 
\[\sum_{k=2}^{n}{n \choose k}{k \choose2}.\]
Hence 
\[{n \choose2}2^{n-2}=\sum_{k=2}^{n}{n \choose k}{k \choose2}.\]
Indeed there are ${n \choose k}$ ways to choose $k$ friends who would go to the party, and we can choose $2$ poeple from the $k$ people who are going to the party in ${k \choose 2}$ ways to bring pizza.  

The answer is accepted.