Combinatoric Proof for $\sum_{k = 1}^{n} k \binom{n}{k} = n2^{n - 1}$

Question

Give a combinatoric proof for:

For $n \in \mathbb N$ $$\sum_{k=1}^{n}k \binom{n}{k} = n2^{n-1}$$

PS: Do not derive the formula by differentiating another formula.

Already answered at http://math.stackexchange.com/questions/1923215/give-a-combinatorial-proof-nn12n-2-sum-k-1nk2-binomnk — Jack D'Aurizio, Mar 02 '17 at 00:09

score 2 · Answer 1 · answered Mar 02 '17 at 00:21

2

We have $n$ people. We want to select some of them, then choose a chief for the selected group (among the chosen people). If we first choose the group, then the chief, we have $\sum_{k=1}^{n}\binom{n}{k}\cdot k$ ways for accomplishing our task. On the other hand, we may also select the chief first, then the other people in his/her group, and this leads to $n 2^{n-1}$ as wanted.

answered Mar 02 '17 at 00:21

Jack D'Aurizio

353,855

1

Thanks! I was confused why we are not over counting. If we have n people to choose the chief from and the remaining n-1 people are either in or out, then wouldn't we also be counting choices like choose kth person as chief and only choose person 1 and 2 not k for the remaining people. I forgot we've already chosen the first person when we chose the chief so it must be included. =) – Xiao Lei Zhang Mar 02 '17 at 01:14

Doug M · Answer 2 · 2017-03-02T00:14:38.147

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$k{n\choose k} = k\frac {n!}{k!(n-k)!} = \frac {n!}{(k-1)!(n-k)!} = \frac {n(n-1)!}{{k-1}!(n-k)!} = n{n-1\choose k-1}$

$\sum_\limits{k=1}^{n}k {n \choose k} = n\sum_\limits{k=0}^{n-1}{n-1 \choose k} = n2^{n-1}$

edited Mar 02 '17 at 00:14

answered Mar 02 '17 at 00:06

Doug M

57,877

You are missing the parentheses in $(k - 1)!$. – N. F. Taussig Mar 02 '17 at 00:13

Combinatoric Proof for $\sum_{k = 1}^{n} k \binom{n}{k} = n2^{n - 1}$

2 Answers2