## InsertCoolNameHere Group Title Ok, I need someone who understands permutations. 2 years ago 2 years ago

True $nCr = \frac{n!}{(n - r)!r!}$ $2C2 = \frac{2!}{(2 - 2)!2!}$ $2C2 = \frac{2!}{0!2!}$ $2C2 = \frac{2!}{2!}$ $2C2 = 1$ $nPr = \frac{n!}{(n - r)!}$ $2P2 = \frac{2!}{(2 - 2)!}$ $2P2 = \frac{2!}{0!}$ $2P2 = 2!$ $2P2 = 2$ 2 is 2 times 1. THis happens because if order is not important and you're choosing 2 from 2, you get 1. It's only natural that if order is important, you can just switch the order of them around and get 2. The same is if it's 2 choose 1.