I'm in a second year discrete mathematics course, and we have identities like this $$\binom{n}{k}(n-k) = \binom{n-1}{k}n$$ and Pascal's Triangle law.
Our professor said that algebraic proofs are fine (and I have them) but is encouraging us to learn combinatorial arguments. I have them for most the rest of the questions; however, this one is stumping me. I'm looking over my notes, and it looks most like $A_k^n$, but that hasn't really gotten my anywhere. I'm not sure how to parse the LHS and RHS into any meaningful counting argument. Any help would be much appreciated!
By combinatorial argument I would understand the challenge to be proving an identity as two ways of counting the same quantity.
Sometimes a bit of lateral thinking can help to spot such an argument. The binomial $\binom{n}{k}$ is a way of counting the $k$-subsets of an $n$-set. Multiplying by $(n-k)$ is how many ways one could single out an element of the remaining $(n-k)$ items left after picking that $k$-subset.
So this is a bit like the "bonus ball" being chosen as the final draw in a lottery. The order of the first $k$ items doesn't matter; they are treated as a $k$-set. But the final ball drawn has a special significance, making it harder to earn the top prize by correctly identifying that number.
If you think of the left hand side in that fashion, it should be easy to find a similar explanation of how the right hand side represents counting the same possible outcomes in a different way. First pick (what), and then pick (the rest of the outcome)?