Why does Binomial Distribution use Combination Formula as Coefficient?

Question

Could someone please help explain why the binomial coefficient is nChoosek when n-k options are not valid to be included in the set of k? I used an example below to illustrate why I am confused on how it works.

Let's take a set of billiard balls. There are 22 different balls and if I wanted to take a set of any 3, non-repeating balls in a combination, I would apply 22Choose3. This states that I have 22 x 21 x 20 valid options, but since it is combination, not a permutation, I divide that by 3!, leaving me with 1,540 valid combinations.

In the case of a binomial coefficient, let's say I have 22 options and I am trying to compute a set of 3 successes. In this case, I do not have 22 x 21 x 20 as the numerator because this suggests each trial was a success and I have 22 successes to choose from for the first option, 21 as the second, and 20 for the third. Therefore if this is the case, then why do we multiply this number of "possibilities" to the probability of success^k*(1-p)^n-k?

Answer 1

You can instead view it as a permutation of $22$ in which $3$ are of one kind, and $19$ of another, which yields

$\dfrac{22!}{3!19!}$, which, of course is the same as $\dbinom{22}3$

Because only two types of results are possible, we can use the binomial coefficient.

If there were three types of results possible, eg win, draw, lose, we would use the multinomial coefficient, viz

$\dbinom{w+d+l}{w,d,l}$, which is the same as $\dfrac{(w+d+l)!}{w!d!l!}$

Answer 2

Here is a simpler example of which the number of outcomes with $k$ successes turns into the binomial coefficient:

Success$=1$, Fail$=0$ where successes occur $p$ of the time. $p^c$ is the complement probability:

From the below graphic, observe that $0$ successes have a $p^3$ chance, $1$ success has a $3p(1-p)^2$ chance of occurring, $2$ successes have a $3p^2(1-p)$ chance, and $3$ successes have a $(1-p)^3$ chance of occurring.

The coefficient is the number of outcomes with $k$ successes at the $3$ trial mark: $\binom{3}{\text{k=number of successes}}=\{1,3,3,1\}$

This pattern implies the general Binomial CDF.