Using the combinational proof template for discrete mathematics is this a valid polished proof?

Question

Let $k, m, n$ be positive integers such that $n\geq k\geq m\geq 0$. Then, ${n\choose k} {k\choose m} = {n\choose m} {{n-m}\choose {k-m}}.$

Here is my Polished Proof:

Claim: Suppose $k$, $m$, and $n$ are positive integers such that $n\geq k\geq m\geq 0$. Then, ${n\choose k} {k\choose m} = {n\choose m} {{n-m}\choose {k-m}}.$

Proof: To prove ${n\choose k} {k\choose m}$ $=$ $ {n\choose m} {{n-m}\choose {k-m}}$ let's imagine that we have a class of $n$ people and we will choose $k$ of them to form a committee, of those on the committee we will choose $m$ of them to be on the subcommittee. We will consider the question: How many $k$-element subsets of people does the set $\{1,2,3...,n \}$ have, and of $k$ people how many, $m$-element subsets of people does $k$ have?

Answer 1: Given $n$ people, we can form a committee of size $k$ in ${n\choose k}$ ways to get $k$-element subsets. Once the committee is selected, we can form a subcommittee of size $m$ in ${k\choose m}$ ways to get $m$-elements. Thus, we can form $k$-element subsets of $n$ and $m$-element subsets of $k$ by ${n\choose k} {k\choose m}$.

Answer 2: On the other hand, we can achieve the same total of ways by forming a subcommittee of size $m$ in ${n\choose m}$ ways to get $m$-element subsets. Once we have selected the subcommittee we can form a committee of the remaining size of $(k-m)$ in ${{n-m}\choose {k-m}}$ to get $(k-m)$-element subsets. We choose $(k-m)$ people from $(n-m)$ people because we have already selected $m$ people for the subcommittee. Thus, we can form $m$-element subsets of $n$ and $(k-m)$-element subsets of $(n-m)$ by ${n\choose m} {{n-m}\choose {k-m}}$.

Since answers 1 and 2 are correct solutions to the same question, ${n\choose k} {k\choose m}$ $=$ $ {n\choose m} {{n-m}\choose {k-m}}$.//

Answer 1

Nice work so far! Some comments:

1. Classes of students don't normally have committees. While it's not required, the metaphor is more convincing when it's familiar.
2. The committee and subcommittee turned into subsets at some point. If you're going to use a metaphor, you should do it consistently.
3. What justifies the multiplication in each answer? That should explained more.