# Combinatorial proof for $\binom{n}{a}\binom{a}{k}\binom{n-a}{b-k} = \binom{n}{b}\binom{b}{k}\binom{n-b}{a-k}$?

I have to prove the following using a combinatorial proof:

$\binom{n}{a}\binom{a}{k}\binom{n-a}{b-k} = \binom{n}{b}\binom{b}{k}\binom{n-b}{a-k}$

Ok, so here is what I have worked out so far:

We have some sets a, b, n, k.

From what I can see in the identity: a is subset of n; k is subset of a; b is subset of n; k is subset of b

Here is what I think the combinatorial proof should be (using the committee forming method):

We have a total of n people. We want to form 2 teams: Team 1 and Team 2, containing a and b number of people, respectively. And elect a total of k people as leaders of the 2 teams.

there are 2 different ways of forming such sets.

Out of n, choose the number of people to be in team 1. $\binom{n}{a}$ ways of doing this. Then we choose to select all the k leaders out of team 1. $\binom{a}{k}$ ways of doing this. Out of the remaining people, select the total number of people to be in team 2. We have already selected a people out of n, and already selected all the k leaders, hence $\binom{n-a}{b-k}$ ways of doing this.

Out of n total people, chose all the people to be in team 2. $\binom{n}{b}$ ways of doing this. Then we choose to elect all the k leaders from team 2. $\binom{b}{k}$ ways of doing this. Out of the remaining (n-b) people, we need to select the people to be in team 1, but since all the leaders are taken from team 2 already, we have $\binom{n-b}{a-k}$ ways of doing this.

What do you guys think?

Most of it makes sense to me, although I am really not sure if I am doing the $\binom{n-a}{b-k}$ and $\binom{n-b}{a-k}$ parts right in each side of the equation.

• Try instead: n = total number of people. a = number of people on team 1 (so n-a people on team 2). b = total number of leaders. k = number of leaders on team 1 (so b-k leaders on team 2). Then the left hand side picks the team then the leaders and the right hand side picks the leaders and then the team. – Bill Cook Oct 2 '11 at 19:08
• @Bill Cook: Ok, let me try it out. – user952949 Oct 2 '11 at 19:16
• @BillCook: you should make that comment into an answer. – Greg Martin Oct 2 '11 at 19:45

Consider this question:

From a group of $n$ people how many ways can we choose $a$ people to have an A on their jersey and $b$ people to have a B on their jersey while having $k$ people with both an A and a B on their jersey?

Once you have explained why both sides must be equal, you should be able to see that both sides must equal $\displaystyle\frac{n!}{(a-k)!(b-k)!k!(n-a-b+k)!}$. To expand my comment:

If you let $n$ be the total number of people, let $a$ be the number of people on team 1 (so $n-a$ people are on team 2), let $b$ be the total number of leaders, and let $k$ be the number of leaders on team 1 (so $b-k$ leaders are on team 2).

Consider the left hand side: $n \choose a$ is the number of ways of selecting team members. Then $a \choose k$ chooses the leaders on team 1. This leaves $b-k$ leaders to be chosen from the remaining $n-a$ people (who are on team 2).

Now the right hand side: $n \choose b$ is the number of ways of choosing leaders. $b \choose k$ is the number of ways of choosing leaders on team 1. Since $k$ people have been chosen for team 1, $a-k$ more members of team 1 must be chosen. These must be chosen from the pool of non-leaders (of which there are $n-b$). The number of ways of choosing the remaining team members is then $n-b \choose a-k$.

Both products are easily seen to be expressions for the quadrinomial coefficient $$\binom n{a-k,\quad k,\quad b-k,\quad n-a-b+k}.$$ In other words they count ways to colour $n$ object with $4$ colours, using each colour repectively $a-k$ $k$, $b-k$ and $n-a-b+k$ times. The different products correspond to two ways of making a first selection into two batches of $2$ colours each first, and then selecting one colour within each batch. See robjohn's anwer for a nice description of two ways of dividing into batches (based on A's respectively on B's). There is no need for the formula for the quadrinomial coefficient.