Combinatorial Proof of ${{n}\choose{k}} {{k}\choose{j}} = {{n}\choose{j}} {{n-j}\choose{k-j}}$ How do I prove by a combinatorial argument that ${{n}\choose{k}} {{k}\choose{j}} = {{n}\choose{j}} {{n-j}\choose{k-j}}$
 A: 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. 
A: 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.
