Yes, that is precisely correct.

(48 choose 47) = (48 choose 1) = 48

This is a combinatorial identity, that is, (N choose K) = (N choose (N-K)).

Do you see why this must be true? Say you've got 48 balls, numbered from 1 to 48. If you choose 47 of them to take away, you can see what you did from a different perspective... that you chose 1 from 48 to leave. It follows then that if you have N differentiable objects, and you choose K of them, it's the same as if you "chose" (N-K) of them to not be chosen.

If you still don't see why this is true, please ask again, and I will try to provide a different example.

RMJ