# Why do we have to use "combinations of n things taken x at a time" when we calculate binomial probabilities?

See below on my thoughts:

#### Explanation:

The general form for a binomial probability is:

sum_(k=0)^(n)C_(n,k)(p)^k((~p)^(n-k))

The question is Why do we need that first term, the combination term?

Let's work an example and then it'll come clear.

Let's look at the binomial probability of flipping a coin 3 times. Let's set getting heads to be $p$ and of not getting heads ~p (both =1/2).

When we go through the summation process, the 4 terms of the summation will equal 1 (in essence, we are finding all the possible outcomes and so the probability of all the outcomes summed up is 1):

${\sum}_{k = 0}^{3} = \textcolor{red}{{C}_{3 , 0} {\left(\frac{1}{2}\right)}^{0} \left({\left(\frac{1}{2}\right)}^{3}\right)} + \textcolor{b l u e}{{C}_{3 , 1} {\left(\frac{1}{2}\right)}^{1} \left({\left(\frac{1}{2}\right)}^{2}\right)} + {C}_{3 , 2} {\left(\frac{1}{2}\right)}^{2} \left({\left(\frac{1}{2}\right)}^{1}\right) + {C}_{3 , 3} {\left(\frac{1}{2}\right)}^{3} \left({\left(\frac{1}{2}\right)}^{0}\right)$

So let's talk about the red term and the blue term.

The red term describes the results of getting 3 tails. There is only 1 way for that to be achieved, and so we have a combination that equals 1.

Note that the last term, the one describing getting all heads, also has a combination that equals 1 because again there is only one way to achieve it.

The blue term describes the results of getting 2 tails and 1 head. There are 3 ways that can happen: TTH, THT, HTT. And so we have a combination that equals 3.

Note that the third term describes getting 1 tails and 2 heads and again there are 3 ways to achieve that and so the combination equals 3.

In fact, in any binomial distribution, we have to find the probability of a single kind of event, such as the probability of achieving 2 heads and 1 tails, and then multiplying it by the number of ways it can be achieved. Since we don't care about the order in which the results are achieved, we use a combination formula (and not, say, a permutation formula).