# Intuition or proof for combinatorial function in binomial distribution

For a the probability of a binomial distribution with n trials, the probability of k successes where Prob(success) = p is (n Choose k) * p^k * (1-p)^(n-k). I understand that we have to multiply p^k * (1-p)^(n-k) by something but what’s 1) an intuitive explanation and/or 2) a proof for why we multiply by the (n Choose k). Especially with regard to the intuitive side of things, I feel tempted to say the factor should be (n Permutation k).

• Multiply by ${n\choose k}$ to get the arrangements of $k$ Successes in $n$ trials, which is found by choosing the $k$ positions for the $k$ Successes. Example: For $X \sim \mathsf{n=3, p=1/2},$ there are $s^3 = 8$ possible outcomes. For $P(X=2)$ the ${3\choose 2} = 3$ relevant outcomes are HHT,HTH, THH and $P(X=2)= 3/8.$ Mar 21, 2021 at 5:59
• @BruceET I think you can convert this to an answer. Mar 21, 2021 at 14:12
• @gunes: OK. Done--with a few extra words. Mar 21, 2021 at 15:58

If $$X \sim \mathsf{Binom}(n,p),$$ then to find $$P(X=k),$$ one multiplies the probability $$p^k(1-p)^{n-k}$$ of a particular outcome with $$k$$ Successes by $${n \choose k},$$ the number of arrangements of $$k$$ Successes among $$n$$ trials, to get the total probability $$P(X=k).$$ This amounts to choosing the $$k$$ positions out of $$n$$ for the $$k$$ Successes.
Example: For $$X\sim\mathsf{Binom}(n=3,p=1/2),$$ there are $$2^3=8$$ possible outcomes altogether, each with probability $$1/8.$$ For $$P(X=2),$$ the $${3 \choose 2} = \frac{3!}{2!\cdot 1!}=3$$ relevant outcomes are HHT,HTH, and THH, so $$P(X=2)=3/8.$$
In R, where dbinom is a binomial PDF (or PMF), this result is found as shown below:
dbinom(2, 3, .5)