Combinatorial argument for the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$ I am looking for the combinatorial argument for the identity:
\begin{equation}
k\binom{n}{k} = n\binom{n-1}{k-1}
\end{equation}
This is easy to show algebraically as:
\begin{equation}
\binom{n}{k} = \dfrac{n(n-1)(n-2)(n-k+1)}{k(k-1)!}
\end{equation}


*

*What is the combinatorial argument?

*What are some general ideas to get started?


Here is a clarification of 2. From what I have seen so far, proving (combinatorially) an identity with an addition sign usually implies that we need to partition a set (this makes sense because of the addition rule and provides a nice visual). On the contrary, the previous observation leads me to believe that multiplication in identities can be resolved with the multiplication principle, but what is the "visual/interpretation" for this? Could someone provide such an interpretation for the example identity given above?
 A: The number of ways of choosing a committee of $k$ members from a group of $n$ people, then selecting one of the $k$ members to be chair, is equal to the number of ways of choosing the chair from out of the whole group of $n$ first, then selecting the $k-1$ other members from among the $n-1$ other people.
A: The LHS is a product, and one principle is that products count pairs. As $\binom{n}{k}$ is the number of $%k$-subsets of $\{1,\ldots,n\}$, we can 
interpret the LHS as the number of ordered pairs $(i,S)$ where $S$ is a subset 
of $\{1,\ldots,n\}$ of size $k$, and $i$ is an element of $S$.
Now we see that we have counted the pairs by choosing first the subset, and then an element from it. If we choose $i$ first, we get a $k$-subset that contains it by choosing a subset of size $k-1$ from $\{1,\ldots,n\}\setminus i$ and adding $i$ to it.
So the number of pairs is $n\times \binom{n-1}{k-1}$. (Here we are making use of a bijection between the $k$-subsets that contain $i$ and the $(k-1)$-subsets of $\{1,\ldots,n\}$ that do not contain $i$.)
