Make sure the combinatorial proof is correct

Question

I am asked to give a combinatorial proof to the identity: ${n\choose7}=\sum_{k=3}^{n-4}{n-k\choose4}{k-1\choose2}$.

Here is my idea: $n\choose7$ implies the number of ways of choosing $7$ elements from a set of $n$ elements. It is equivalent to partitioning the $n$-set into a $(n-k)$ set, a $(k-1)$ set and an $1$ set. From $(n-k)$-set and $(k-1)$-set we choose $4$ and $2$ elements respectively. In order to exhaust all possible combinations we let $k$ run from $3$ to $n-4$, which is the above identity.

Is my idea on the right way?

Answer 1

Yes, you are on the right track. The $1$-set contains only the third smallest element $k$ of the $7$-set. With this correspondence, each $7$-set gets counted exactly once in the sum.

Answer 2

Using stars & bars and product rule. For a given $k$, there are $\binom{n-k}{4}\binom{k-1}{2}$ positive integer solutions to the following equations.

$$ \sum_{i=1}^{5}a_{i}=n+1-k \phantom{xx} \text{and} \phantom{xx}\sum_{i=6}^{8}a_{i}=k $$

If we sum the number of solutions for all possible $k$, we will obtain the number of ALL positive integer solutions to the following equation:

$$ \sum_{i=1}^{8}a_{i}=n+1 $$

And that number is $\binom{n}{7}$ (stars & bars again).