TITLE: Is there a story proof behind the combinatorial identity $(n-2k)\binom{n}{k} = n\left[ \binom{n-1}{k} - \binom{n-1}{k-1} \right]$?
QUESTION [6 upvotes]: Is there a "story proof"/combinatorial proof for the following combinatorial identity:
$$(n-2k)\binom{n}{k} = n\left[ \binom{n-1}{k} - \binom{n-1}{k-1} \right]\tag1$$
I know that this identity can be proved by using the following identities:
$$k\binom{n-1}{k} = (n-k)\binom{n-1}{k-1}\tag2$$
$$k\binom{n}{k} = n\binom{n-1}{k-1}\tag3$$
but is there a "story proof" for equation $(1)$?
Edit 1: I do know the story proofs for equations 2 and 3. But 'sewing them together' is the problem!
$$\text{RHS} \stackrel{\text{i}}{=}  n\left[ \binom{n-1}{k} - \binom{n-1}{k-1} \right] \stackrel{\text{ii}}{=}  \frac{n}{k}\left[ k\binom{n-1}{k} - k\binom{n-1}{k-1} \right] \stackrel{\text{iii}}{=} \frac{n}{k}\left[ (n-k)\binom{n-1}{k-1} - k\binom{n-1}{k-1} \right] \stackrel{\text{iv}}{=} \frac{n}{k}\binom{n-1}{k-1}\left[ (n-k) - k \right] \stackrel{\text{v}}{=} (n-2k)\binom{n}{k}$$

Precisely, how do you formulate a story proof for step (iv)? i mean the term $\binom{n-1}{k-1}$ is being taken common in step iv. What could a story proof for taking a term common be?

REPLY [3 votes]: Following is a nice combinatorial proof for the identity as written:

Consider a class of $n$ students. You want to reward $k$ number of students, by taking money from the remaining $n-k$ students. What is the money that you will have in the end?

We choose $k$ students to be rewarded in $\dbinom nk$ ways. Money you'll get is $n-k$ dollars, from the remaining students, and the money you'll pay is $k$ dollars. So, you'll have saved $(n-2k)\dbinom nk$ dollars in the end.
This can also be calculated by considering a particular student (to be chosen in $n$ ways). He will pay every time when he is in the non-rewarded set, and he'll be in the non-rewarded set $\dbinom{n-1}k$ number of times, when the rewarded set is chosen by excluding him. He'll get paid every time he is in the rewarded set, and he will be in rewarded set $\dbinom{n-1}{k-1}$ number of times. Therefore, the total money he pays is $\dbinom{n-1}k-\dbinom{n-1}{k-1}$. Multiplying this by $n$ gives the net amount paid by the class, and we are done.