Prove in a combinatorial way (for example using an argument with committees) for the Identity

Question

$$\sum^{n}_{k=2}k(k-1)\binom{n}{k} =n(n-1)2^{n-2}$$

$$n \geq 2$$

I'm recently new learning combinatorics and having trouble understand the intuition behind this. I don't either know what's combinatorial argument with committees.

Answer 1

Assume you are the headmaster in a school. You have to choose from $n$ teachers which ones are going to teach Class 1A. Out of them, one has to be the headteacher, and one has to be the deputy headteacher. How many amounts of ways can you make the choices?

One way to count is this: you'd take $k$ teachers who will teach the class, one out of $k$ could be the headmaster, and one out of the remaining $k-1$ could be the deputy headmaster. This goes from $k=2$ to $k=n$, thus the sum is: $$\sum^{n}_{k=2}k(k-1)\binom{n}{k}$$ Now we count it from another way. From the $n$ teachers, we'll choose one as the headteacher, and one out of the remaining $n-1$ as the deputy headteacher. The remaining $n-2$ have two choices: they are either teaching the class or not. $n(n-1)$ ways of choosing the head- and deputy headmaster, and $2^{n-2}$ choices for picking remaining teachers for the class. This will equal to $$=n(n-1)2^{n-2}$$ choices, and since we are counting the same thing, the two expressions should be equal.