# Why binomial distribution doesn't count permutations?

Why in Binomial distribution the formula starts with $n\choose k$ and not with something like $k!\over n!$? Isn't the order important? Or, it is important but due to the independence of the event?

• $k!n!\over n!$ would be the same as $k!$...as you sure that's what you meant to write? Commented Jul 10, 2015 at 13:26
• Do you know what $\binom nk$ represents ?
– user65203
Commented Jul 10, 2015 at 13:34
• @man_in_green_shirt right, I mean k fixed points in n permutations
– gbox
Commented Jul 10, 2015 at 13:34
• The binomial distribution is also order independent: you don't specify any order, do you ?
– user65203
Commented Jul 10, 2015 at 13:39
• I recommend you to reconstruct binomial distribution trough an example forgetting absolutely all that you read in some book. By example count, graphically, the probability to get a 6 in n throws of a fair dice of six sides. You only need to know that probability is $\frac{\text{number of cases that I want to know it probability}}{\text{all cases}}$ Commented Jul 10, 2015 at 15:55

The binomial distribution is frequently used to model the number of successes in a sample of size $n$ drawn with replacement from a population of size $N$.

Note two things:

• What is key is the number of successes. So if with four tries you have $(S,F,F,S)$ or $(F,S,F,S)$ is irrelevant, in both cases you have two successes.

• Since one draws with replacement each try is independent from what happened before. Thus $(S,F,F,S)$ and $(F,S,F,S)$ certainly have the same probability.

And, $\binom{n}{k}$ is the number of ways in which you can have $k$ successes among $n$ tries.

If you do not draw with replacement, but without replacement, then the second point is no longer true and you get a different distribution, called hyper-geometric distribution.

• No. It is $\binom{2}{1}p (1-p)$ Generally it is $\binom{n}{k}p^k (1-p)^{n-k}$ where $n$ number of tries, $k$ number of successes, and $p$ probability of success.
– quid
Commented Jul 10, 2015 at 15:38
• Sorry you respond very quick I will put it that way: If I want 1 "win" out of 6 tries, it turn out that it does matter in which try I won (1,0,0,0,0,0) or (0,1,0,0,0,0) etc. doesn't it mean that there is importance to the order? (unlike $n\choose k$ that does not has importance to order)
– gbox
Commented Jul 10, 2015 at 15:46
• Yes, they are two case, and you want to take both into account, together with all the other cases with two S and two F which are exactly $\binom{4}{2}$. The binomial coefficient $n$ over $k$ is there precisely to count the number of cases with $k$ wins in $n$ tries. You want to take all of them together, so you need to know how many there are.
– quid
Commented Jul 11, 2015 at 17:58
• Not really. You have the following possibilities with two success SSFF, SFSF, SFFS, FSSF, FSFS, FFSS. These are all different, but you only care about the number of successes. So you do not care which one of those happened. The prop for one particular of them to happen is $p^2(1-p)^2$ so you have $6 p^2(1-p)^2$ for one of the $6$ to happen. And the $6$ is just the $4$ over $2$.
– quid
Commented Jul 11, 2015 at 18:16
• Let me give you still another example. You toss a coin three times, the possibilties are TTT, TTH, THT, HTT, THH, HTH, HHT, HHH. these are all posibilties. Each event has a prob of 1/8. Now it depends what you want to know. If you want to know the prob that you have H exactly once it is 3/8 as there are three different ways to get this. Precisely since you do not care which of the three it was, you have to multiply by the number of possibilties. By contrast the prob for HTT is 1/8.
– quid
Commented Jul 11, 2015 at 19:17

$$\frac{n!}{n_1!n_2!},\;\; n_1+ n_2=n,$$

counts the permutations of the sequence $$\underbrace{p_1...p_1}_{n_1}\underbrace{p_2...p_2}_{n_2}.$$

Coincidentally, $$\frac{n!}{n_1!n_2!}=\frac{n!}{k!(n-k)!}={n \choose k}, \;\; n_1=k=\text{number of successes}.$$

Therefore, ${n \choose k}$, disguised as counting combinations, actually counts permutations.

• Thats a cool perspective.
– kaka
Commented Jul 10, 2015 at 15:22