$$\binom{k}{i}\binom{n}{k}=\binom{n}{i}\binom{n-i}{k-i}$$
This identity could be easily shown using algebraic formula of combination. However, I would like to provide a combinatorial proof.
I considered applying Pascal's equality, but got stuck.
Any advice ?
We count the number of ways a committee of $k$ people with a subcommittee of $i$ people can be selected from $n$ people in two different ways.
The left-hand side counts the number of ways of selecting a committee of $k$ people from $n$ available people, then choosing $i$ of those $k$ people to serve on a subcommittee.
The right-hand side counts the number of ways of selecting the subcommittee of $i$ people from the $n$ available people, then choosing the remaining $k - i$ members of the committee from the remaining $n - i$ available people.
We can say , $$\binom{n}{k} $$ Is the no. of ways we can choose $k$ things from $n$ diff. objects, $$\binom{k}{i}$$ Is the no. of ways of choosing $i$ things from $k$ diff. objects,
So this is no. of ways of first choosing $k$ objects from $n$,and then again choosing $i$ objects from $k$.
But we could do the other way round,we could first choose $i$ objects from $n$ ,and then choose $k-i$ objects from remaining $n-i$
We could write the no. ways through this method to be, $$\binom{n}{i}\binom{n-i}{k-i}$$
