Find a combinatorial proof to 10!=7!6!

Prove combinatorically that $ 10!=7!6! $.

By "prove combinatorically" i mean everything that does not pass by computing $10!,7!,6! $ or some factors of them. This is clearly equivalent to find a bijection (or prove it does exist) between $S_{10}$ and $ S_7 \times S_6 $ where $S_n$ is the symmetric group of order $n$.

Here is a combinatorial proof:

Suppose we have ten people and we want to find how many ways to arrange them in a straight line. The answer is obviously $10!$, because we can choose from ten people for the first spot, nine for the second, and so on.

Now consider choosing an order in a different way. Let us put $7$ people in a line, then insert the remaining $3$. There are $7!$ ways to make the $7$-person line, and then there are $8\cdot9\cdot10$ ways to insert the last three people. Thus the total number of ways is $$7!\cdot8\cdot9\cdot10$$ and $$10!=7!\cdot8\cdot9\cdot10$$ Now all you need to do is prove that $\cdot8\cdot9\cdot10=6!$. Can you do that by following my example for the first half of the problem?