Prove that the number of ways to choose two things from a set of $n$ things is $n(n+1)/2$

Question

In other words, prove $${n \choose 2} =\frac{n(n+1)}{2}$$

I am attempting to solve this proof using combinatorics. I know that by definition, $${n \choose 2} = \frac{n!}{2!(n-2)!}$$

So equivalently $$\frac{n(n+1)}{2}=\frac{n!}{2(n-2)!}$$

$$\frac{n(n+1)}{2}=\frac{n(n-1)(n-2)!}{2(n-2)!}$$

$$\frac{n(n+1)}{2}=\frac{n(n-1)}{2}$$

But how is this possible? I appreciate any input–thanks in advance!

Answer 1

The answer is indeed

$$ \binom{n}{2}=\frac{n(n-1)}{2}. $$

You actually proved this above. There must be a misprint in the original question.

Otherwise, as you concluded, $$ \frac{n(n+1)}{2}=\frac{n(n-1)}{2}\implies n=0. $$