How many possibilities for a $12$ digit binary number with six $1$s and six $0$s?

Question

I am trying to figure out how many possibilities there is to build a $12$-digit binary number with six $1$s and six $0$s.

One example would be: $111111000000$

Answer 1

Hint: The leading digit must be a $1$, for otherwise the number would have fewer than $12$ digits. That leaves you with eleven positions to fill with five $1$s and six $0$s. The number is completely determined by choosing which five of the remaining $11$ positions will be filled with $1$s. The number of ways a subset of $k$ elements can be selected from a set with $n$ elements is $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$ where $n!$ is the product of the first $n$ positive integers if $n$ is positive and $0! = 1$.