Combinatorial Proofs: Show that $\binom{n}{j+1}(j+1) = \binom{n}{n-j}(n-j)$

Question

I'm not sure if I understand why the following proof holds and I'm not sure where to start:

$$ {n \choose j+1 } (j+1) = {n \choose n-j} (n-j) $$

Would appreciate any help and why this does hold (in particular the movements that allow the switches to happen... like does just saying j+1 = n-j work?)

Edit: Sorry, forgot to mention that I cannot use the identity: $$ {n \choose k} = \frac{n!}{k!(n-k)!} $$

Answer 1

Both sides count, for example, the number of ways of building a team of $j$ regular players and a captain from a roster of $n$ available athletes.

The left hand side can be interpreted as picking a team of $j+1$ players from $n$ available, and then selecting a team captain from among those players: the $\binom{n}{j+1}$ picks the players, and then you have $j+1$ ways to pick the captain.

The right hand side can be interpreted as first picking everyone who will not be a regular player in the team (by picking $n-j$ people who won't be players), and then picking from among them who will be the captain (you have $n-j$ ways of doing that).

In both cases, you end up picking a team with $j$ regular players and a captain from a roster of $n$ potential athletes.