Present a combinatorial argument for the identiy $\sum^{n}_{k=1} k\binom{n}{k} = n\cdot 2^{n-1}$ This is a question in my textbook that does not provide a solution. Any help on a solution?
Consider the following identity:
$\sum^{n}_{k=1} k\binom{n}{k} = n\cdot 2^{n-1}$
Present  a combinatorial argument for the identity by considering a set of $n$ people and determining,in two methods, the number of ways you can select a committee of any size and a chair personfor the committee 
(a) How many ways possible you can select a committee of size k and its chairperson?
(b) How many ways possible you can select a chairperson and the other committee members?
 A: Okay. We can form a committee with a chairperson of size $k$ by first choosing $k$ people from our group of $n$, and from those $k$ people we may choose someone to chair the committee. There are $\binom{n}{k}$ ways to perform the first task, and $k$ ways to perform the second task, and so $k\binom{n}{k}$ ways to form a committee of size $k$ with a chairperson. 
Summing over $1\le k\le n$, we have the number of ways to form a committee with a chairperson of size less than or equal to $n$ is precisely $\sum_{k=1}^{n}k\binom{n}{k}$. 
You can also form a committee with a chairperson of size less than or equal to $n$ by choosing the chairperson first, then the rest of the committee. There are $n$ choices to choose the chairperson from. For the remaining $n-1$ people, the selector has two choices for each person: to include them or not. So we have $n$ choices for the chairperson, $2$ choices for the next, $2$ choices for the next, etc. Multiplying these together, we have $n2^{n-1}$ ways to form such a committee, which proves the identity. 
A: Given a group of $n$ people, we can count the number of ways to choose a committee of some size (at least one member) from the group, and choose its leader. If we count this in two different ways, our results must be equal.

First count:
We count the committees by size $k$, from $k=1$ to $k=n$. For each $k$, there are $\binom{n}{k}$ ways to choose which of the $n$ people is in the committee. After choosing the members, there are $k$ ways to choose their leader from them. This gives $$\displaystyle\sum\limits_{k=1}^n k \binom{n}{k}$$

Second count:
We first choose the leader of the group, since there must be one. There are $n$ choices for this leader. Then, we decide which of the other $n-1$ people are on the committee (all as non-leaders). For each of the other $n-1$ people, we have $2$ choices: member or non-member. So there are $2^{n-1}$ choices for the other committee members, for a total of $$n \cdot 2^{n-1}$$

Thus, $$\displaystyle\sum\limits_{k=1}^n k \binom{n}{k} = n \cdot 2^{n-1}$$
