Combinatorial Proof of ${{n}\choose{k}} {{k}\choose{j}} = {{n}\choose{j}} {{n-j}\choose{k-j}}$

Question

How do I prove by a combinatorial argument that ${{n}\choose{k}} {{k}\choose{j}} = {{n}\choose{j}} {{n-j}\choose{k-j}}$

Answer 1

We can proceed this way

For LHS, We are to permit $k$ people inside bus from a queue of $n$ people. And then choice is made for $j$ of those $k$ are provided a seat.

For RHS, We are to choose at first $j$ people from $n$ who can sit. Now from remaining $n-j$ people we choose $k-j$ people who will board without seat. Thus both cases are same.

Answer 2

Lets have this combinatorial problem: "Find out how many ways you can distribute $j$ apples and $k-j$ oranges among $n$ persons, when your task is to give anyone one piece of fruit at most."

One way to do that is to pick $k$ people, who will receive some fruit and of these people choose $j$ people, that will get apples. The rest of $k-j$ people will logically get oranges in only one way possible. That leads us to the number on the left side.

Other way to do that is to pick $j$ people that will get the apples, and from the rest of $n-j$ people, you can pick $k-j$ ways the rest, that will get the oranges". That gets us the number on the right side.

Both ways solve the same problem, so the numbers must be the same.