Combinatorial argument for a homework question: $\binom nr \binom r3=\binom n3\binom{n-3}{r-3}$

Question

$${n\choose{r}}{r\choose3}={n\choose{3}}{{n-3}\choose{r-3}}$$ where $n\ge r\ge3$

I was able to show this algebraically using the combinations formula but I'm unsure how to approach it using the combinatorial argument.

So far I've gotten this:

There are $n\choose{r}$ ways to choose subsets of $r$ elements from a set of $n$ elements

There are $r\choose3$ ways to choose subsets of 3 elements from a set of $r$ elements

Therefore, by the product rule, there are $n\choose{r}$$r\choose3$ distinct ways to choose $r$ elements from a set of $n$ elements then to choose 3 elements from a set of $r$ elements.

At this point, I fail to see the connection between the two sides of the equation.

For future reference: write entire mathematical expressions, not just individual symbols, in math mode. In other words, write $f(x)=\frac{1}{x} + 10$ (which yields "$f(x)=\frac{1}{x} + 10$") instead of f(x)=$\frac{1}{x}$ + 10 which yields a much uglier "f(x)=$\frac{1}{x}$ + 10" — 5xum, Oct 19 '21 at 06:44
See also: Combinatorial proof of $\binom{k}{i}\binom{n}{k}=\binom{n}{i}\binom{n-i}{k-i}$ and Combinatorial Proof of ${{n}\choose{k}} {{k}\choose{j}} = {{n}\choose{j}} {{n-j}\choose{k-j}}$. — Martin Sleziak, Oct 19 '21 at 06:51

score 2 · Accepted Answer · answered Oct 19 '21 at 06:54

You can view it as:

I have $n$ people in my team, and I want to choose $r$ of them to be in the exco committee. Out of these $r$ exco committee members, three of them shall be the committee directors.

For LHS, we first have $\binom{n}{r}$ ways to choose $r$ members to be in the exco committee, followed by $\binom{r}{3}$ ways to choose $3$ of these members to be directors.

For RHS, we can instead choose $3$ members in the team to be the directors immediately. This leaves $n - 3$ non-director members, and there are $\binom{n-3}{r-3}$ ways to choose $r - 3$ non-director exco committee members to fill up the remaining committee.

score 0 · Answer 2 · answered Oct 19 '21 at 06:54

I shall use an analogy to explain the combinatorial argument to make things a bit easier.

Imagine you have $n$ players and you need to choose $r$ from them to make a team. Additionally you also have to choose $3$ captains for the team from the $r$ players that you choose.

This can be represented directly by your LHS. Choose $r$ from $n$, and then choose $3$ from $r$.

Alternately, you can choose the 3 captains first from the n players available, and then pick out the rest of the $(r-3)$ players from the remaining $(n-3)$ players not picked yet.

This can be represented by your RHS. Choose $3$ from $n$, and then choose $r-3$ from $n-3$.

I hope you that you can now frame your combinatorial argument.

Combinatorial argument for a homework question: $\binom nr \binom r3=\binom n3\binom{n-3}{r-3}$

2 Answers2