Double counting $n2^{n-1} = \sum_{k=0}^{n} k\binom{n}{k}$

Question

I am trying to prove by double-counting that

$$k2^{k-1} = \sum_{i=0}^{k} i\binom{k}{i}$$

where $n$ is a positive integer.

I did it algebraically. However, the professor told me that I have to do this also by double-counting but I am not sure how to do this. Can someone help me?

Does this answer your question? How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$? You can see many answers here, including by expansions, by double-counting elements of a set, by algebraic manipulation, and the entire lot. — Sarvesh Ravichandran Iyer, Feb 02 '21 at 12:10
I think this answer of the cited question is what OP is after. — qualcuno, Feb 02 '21 at 12:13
See also: Combinatorial proof of $\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$. — Martin Sleziak, Sep 30 '21 at 14:16

score 1 · Answer 1 · answered Feb 02 '21 at 12:18

Consider the problem of forming a committee with a president from a group of $n$ people.

On one hand, you can first fix a president ($n$ choices) and then choose a subset of the remaining people to form the committee ($2^{n-1}$ choices). This gives you LHS.

On the other, you can fix a size $k$ (between $1$ and $n$) first and then pick a committee with $k$ people ($\binom{n}{k}$ many choices) and then pick a president from that ($k$ choices). Sum over all possible $k$ to get the RHS.

Double counting $n2^{n-1} = \sum_{k=0}^{n} k\binom{n}{k}$

1 Answers1