# Prove the number of combinations is equal to $\frac{n!}{(n-j)!j!}$

Considering $n$ distinct elements, and an ordered set of $j$ elements (without repetitions and in which the order doesn't matter); Example: $(a,b,c)$, $n=3$, $j=2$, there are 3 arrangements possible; $(ab,ac,bc)$

I have to prove that the number of combinations is equal to $$\frac{n!}{(n-j)!j!}$$

How do I do that?

Pick $j$ elements. For the first one, you can choose among $n$ possibilities. For the second one, among $n-1$, and so on. Thus in total there are

$$n(n-1)(n-2)\cdots (j+1)=\frac{n!}{(n-j)!}$$ possible drawings.

But these contain all permutations of all the subsets. As order doesn't matter, there are $j!$ replicas.

Hence the formula

$$\frac{n!}{(n-j)!j!}.$$

With your example, you can draw in $3\cdot2=3!/1!$ ways

$$ab,ac,ba,bc,ca,cb$$

and each unordered drawing appears twice.