# N choose K (nCk) usage?

1. Jun 30, 2013

### sherrellbc

1. The problem statement, all variables and given/known data
The binomial coefficient formula can be used in many applications. What I have been learning about is its application in Binomial Distributions.

Define a random variable X to be the number of heads associated with each flip.

Imagine you have, for example, 5 coins. What is the probability that X = 2?
That is, P(X=2)

So, below I have prove how you can use the nCk formula for situations like this one in which you have 1 of two possible outcomes. But what if you were to have many more? Imagine instead of H/T, you have A,B,C,D. How can we apply the nCk in situations like this?

2. Relevant equations

${(n k)} = {n!}/{k!(n-k)!}$

3. The attempt at a solution
Without applying the formula, it is easy to derive(essentially the formula in the process) the solution to the problem.

Given 5 "spaces", the first H has 5 potential slots, then for the next H there remains 4. That is,

_ _ _ _ _
5 4

Which is 5*4 and can be written as, 5!/(5-2)!
Then we want to divide out by all possible permutations of 2 "things", in this case heads.

So, we have:
${(5 2)} = {5!}/{2!(5-2)!} = 10*{1/2^6}$

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But, what if we had more possible outcomes? (i.e. A, B, C ... )
The use of nCk in the previous manner worked because we "knew" what the other spaces would hold. That is,

_ _ _ _ _
5 4

We know that the empty spaces here WOULD be tails, since we defined Heads as success.

For example:
H H T T T
H T H H H
H T T H H
H T T T H
...

What if we used the same logic in the case where we have many more possible outcomes?(1, B, C ... )

Redefine X to be: Number of A's
Find P(X=2)
Using the same logic, we resolve the same solution! That:
${(5 2)} = {5!}/{2!(5-2)!} = 10$
That only difference being that we have no information on probability of A, B, C happening as we did with the Heads/Tails flips.

So, we are essentially saying here that we have 10 UNIQUE combinations of two A's with any other selection of B, C, D .. ect.

How is this possible?

tl;dr, same result with more possible outcomes does not make immediate sense.

2. Jun 30, 2013

### sherrellbc

I do not wish to edit the above post because each time I do all of the formatting quits working.

So, to clarify:
We resolve a solution of 10 unique combinations of in 5 spaces consisting of to A's.

My question is essentially how do we deal how many possible outcomes exist? By this I mean, H/T or A/B/C/D .. etc.

We could have only 6 spaces, but an infinite number of outcomes in the form of A,B,C .. and still resolve 10 in this problem!

What am I not getting here? The binomial application cannot be restricted in this sense to cases in which there are only two possible outcomes? I thought that idea was encapsulated in the Bernoulli Distribution.

3. Jun 30, 2013

### awkward

4. Jun 30, 2013

### sherrellbc

Perfect, so the Binomial Distribution is for only two outcomes. If so, that makes perfect sense because this was not.

If the Binomial Distribution, in general, is for only two possible outcomes, what is the purpose of the Bernoulli Distribution which is a special case of the Binomial?

I will have to read more on the topic it seems. Thank you.

5. Jun 30, 2013

### LCKurtz

You want the multinomial distribution. Look here for one explanation, or use Google for others:

http://onlinestatbook.com/2/probability/multinomial.html

 Woops, I didn't see the other posts...