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$n(n+1)2^{n-2} = \sum_{i=1}^ni^2\binom{n}{i}$

I would like to prove above identity combinatorially.

$n(n+1)$ of LHS is doubled amount of summing up from $1$ to $n$ and remaining $2^{n-1}$ denotes that there's some $n -1$ consecutive choice of something being exist or not.

How could I relate this two factors into potential those of RHS ?

Or will there any other option that I can refer to?

Any advice would be appreciate.

Martin Sleziak
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Beverlie
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2 Answers2

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Hint:

The right-hand side counts the number of ways to choose a subcommittee (of any non-empty size) from a committee of $n$ people, choose a leader, and choose a note-taker... with the understanding that one person can be both the leader and the note-taker.

Can you see how to show that the left-hand side counts the same? It might help to note that $n+1=(n-1)+2$.

Nick Peterson
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  • Hi! Just a by-stander here enjoying the problem and your hint! I was wondering if the method to show that the LHS is indeed equal to the number of subcommittees described by your hint is through induction (which I feel it may be), or a more direct method? – Jihoon Kang Sep 22 '17 at 02:49
  • @Nick Peterson To which point of your explanation the $i^2$of RHS being referred to ? – Beverlie Sep 22 '17 at 02:54
  • @Beverlie In Nick Peterson's answer, for each subcommittee of size $i$, there are $i^2$ ways to designate a leader and a note-taker if we allow the case of one person being both – Jihoon Kang Sep 22 '17 at 02:58
  • @JihoonKang Thanks so we are just opening two positions only! I thought ... intends to say more than 2 positions. – Beverlie Sep 22 '17 at 03:00
  • @Beverlie ah! Fair enough. I am still stuck on showing that the LHS is indeed the number of such subcommittees but I shall work on it so long as I am on this train and have time to think! – Jihoon Kang Sep 22 '17 at 03:03
  • @JihoonKang check out my below answer please – Beverlie Sep 22 '17 at 03:37
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Following Nick's hint I'd like to construct the RHS.

First ask each one of n people whether he/she wants to participate in sub-committee $ = 2^n$

Then I choose $2$ people among the entire $n$ to designate each one of $2$ into the leader or note-taker position = $\binom{n}{2}\cdot2$

but the possibility is only $1\over2^2$ which I successfully choose 2 people who both of them said "YES! I want to be a member of sub-committee with extra work!" =$\binom{n}{2}\cdot2\cdot{1\over4}$

Now, I want to consider $2$ positions only filled with one person. It's simple. Just take $1$ among $n$ with $1\over2$ of possibility = $n\over2$

Now, summing up, the total counts will be $2^n({n(n-1)\over4}+{n\over2})$ which is equal to LHS

Beverlie
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  • Thank you :) this seems to make sense! – Jihoon Kang Sep 22 '17 at 05:02
  • Alternatively, if we choose two different people to occupy the two positions, there are $n$ ways to choose the committee head, $n - 1$ ways to choose the scribe, and $2^{n - 2}$ ways to choose the remaining members of the committee from the remaining $n - 2$ people. If we choose one person to serve as both committee head and scribe, there are $n$ ways to choose that person and $2^{n - 1}$ ways to choose the remaining members of the committee from the remaining $n - 1$ people. Hence, the number of committees is $$n(n - 1)2^{n - 2} + n2^{n - 1} = n2^{n - 2}(n - 1 + 2) = n(n + 1)2^{n - 1}$$ – N. F. Taussig Sep 22 '17 at 08:49