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I want to show $$\sum_{k=1}^{n}k\binom{n}{k} = n2^{n-1}$$ by way of combinatorial proof.

Here is my attempt: Let there be $n$ people. We will count the number of ways in which we can choose a committee of any size and its chairperson, in two ways.

First, we will find a committee and then find a chairperson for that committee. Let the size of the committee be $k\leq n.$ There are $\binom{n}{k}$ of choosing this committee. Then, from the $k$ members chosen, there will be one chairperson, which gives us $k$ choices for the chairperson. By the basic principle of counting, there are $$k\binom{n}{k}$$ possible choices for a committee and its chairperson to be chosen (of size $k$). Since $k$ can be any size from 1 to $n$, with each size mutually exclusive, there are $$\sum_{k=1}^{n}k\binom{n}{k}$$ choices for committees and their chairpersons.

The second way by which we count is to first find the chairperson and then find the rest of the members of the committee. Again, denote the size of the committee by $k.$ We first choose a chairperson, having $n$ choices for the same. After choosing the chairperson, we need $k-1$ members from the remaining $n-1$ members (other than the chairperson since only one member can be the chairperson).

This gives $$n\binom{n-1}{k-1}$$ choices for a committee of size $k.$ Note that since the chairperson is present, there is at least one member, so $k\geq 1.$ Now, to find all choices, we note that the sizes are mutually exclusive and range from 1 to $n$ giving $$\sum_{k=1}^{n}{n}\binom{n-1}{k-1} = n\sum_{k=0}^{n-1}\binom{n-1}{k} = n2^{n-1}.$$

This shows $$\sum_{k=1}^{n}k\binom{n}{k} = n2^{n-1}$$.

Please see if my proof is correct and there are ways to improve it.

Seeker
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Chirag Kar
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  • The suggested answer has a proof strategy which I have followed here as well. The purpose of asking this question is to know whether my understanding of the method of Combinatorial proof is correct. – Chirag Kar Jul 29 '23 at 12:07
  • This first part of your question was already answered in the 1st sentence below, and the second part ("ways to improve it") can be found in that duplicate. – Anne Bauval Jul 29 '23 at 20:40

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Your proof is correct. I would count the left-hand side in the same way you did. For the right-hand side, there are $n$ ways to select the chairperson. The remaining members of the committee are selected from the remaining $n - 1$ people. The number of ways we can select a subset of those $n - 1$ people to serve on the committee is the number of subsets of $n - 1$ people, which is $2^{n - 1}$. Hence, there are $n2^{n - 1}$ ways to select a committee with a chairperson from a set of $n$ people.

N. F. Taussig
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    Ask each non-chairperson whether they want to join the committee or not. There are $n-1$ people, each giving a yes-or-no answer. It is not difficult to see that there are $2^{n-1}$ different possible constellations. – Arthur Jul 29 '23 at 10:38
  • I see, I justified the same by summing it over all possible sizes of the committees. The justification is given by Arthur is also quite interesting, something I did not think about initially. – Chirag Kar Jul 29 '23 at 12:03