Binomial Combinations Proof

Question

I know that the number of unique combinations of $n$ $1$s and $0$s with $x$ $1$s, $n-x$ 0s is:

$${n \choose x} = \frac{n!}{x!(n-x)!}$$

What's the proof? I've looked around and can't find anything.

Thanks

Answer 1

Line up $n$ people. Pick $x$ of them to be on Team 1, and the remaining $n-x$ people are on Team 0.

There are $n!$ ways to line up the $n$ people.

However, you have to divide by the number of ways to line up the people on Team 1, so that you don't double-count any of the teams. There are $x!$ ways to line up the Team 1 players in their places within the big line.

Likewise, there are $(n-x)!$ ways to line up the Team 0 players.

Hence, the number of unique combinations is

$$_nC_x = \frac{n!}{x!(n-x)!},$$

also known as "$n$ choose $x$."

Answer 2

If no replacement is allowed, you can choose the first element among $n$, the second among $n-1$, the third among $n-2$... hence in

$$n(n-1)(n-2)\cdots(n-x+1)=\frac{n!}{(n-x)!}$$ ways. This is called the number of arrangements of $n$ items taken $x$ by $x$.

Because of order, all selections are repeated $x!$ times (with permutations). Hence if order does not matter,

$$\binom nx=\frac{n!}{(n-x)x!}.$$