Prove using combinatorics $\sum\limits_{r=0}^n r^2\binom{n}{r}=n(n+1)*2^{n-2}$.

Question

The left side is choosing $r$ persons from $n$ persons and make one leader and one co-leader such that the leader and the co-leader can be the same.But then the left side should be choose a leader and co-leader and then choose the other which makes the left side $n^2*2^{n-2}$ where did I make a mistake?

If leader and co-leader are the same person, you get $2^{n-1}$ ways to build your team — Exodd, Apr 02 '17 at 16:15

score 2 · Accepted Answer · answered Apr 02 '17 at 16:24

2

It's easier if you rewrite the right side as $n(n-1)2^{n-2} + n2^{n-1}$, then count the two cases separately (leader and co-leader are same, or different.)

answered Apr 02 '17 at 16:24

Thomas Andrews

177,126

Aaditya Vaze · Answer 2 · 2017-04-02T17:12:58.103

The reasoning for the right hand side is wrong, when you are choosing the leader and then the co-leader you are considering it as n^2 cases and then for choosing the other members you are considering 2^(n-2) ways, however if the leader and coleader are same then there are 2^(n-1) ways to choose the team and if they are different then there are 2^(n-2) ways to choose the team, so you will have to break it up into two cases as suggested by @Thomas Andrews

Prove using combinatorics $\sum\limits_{r=0}^n r^2\binom{n}{r}=n(n+1)*2^{n-2}$.

2 Answers2