Binomial coefficient

The binomial coefficient $\binom{n}{k}$ (read "n choose k") counts how many subsets of size $k$ can be drawn from $n$ distinct items, which is the same as counting how many arrangements of $k$ successes and $n - k$ failures fit into a sequence of $n$ trials. The formula is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, with $0!$ defined to equal 1 and $k$ a whole number from 0 to $n$.

Take 5 free throws and ask in how many ways exactly 2 of them could be the makes: $\binom{5}{2} = \frac{5!}{2!\,3!} = \frac{120}{2(6)} = 10$. Those 10 orders are what the coefficient contributes to a probability. If the shooter makes 70 percent of her attempts, the chance of exactly 2 makes is $10(0.7)^2(0.3)^3 = 10(0.49)(0.027) = 0.1323$.

The error to name is treating the coefficient as a probability. "$\binom{10}{5} = 252$, so the probability is 252" is wrong twice over: 252 is a count of orders, and no probability can exceed 1. Nothing about chance enters until that count is multiplied by $p^k(1-p)^{n-k}$, which is the probability of one particular order.

The second slip is using the ordered count instead. Picking 2 of 5 shots in order gives $5 \times 4 = 20$ possibilities, but that counts "first and third" separately from "third and first" even though the pair of makes is identical. Dividing by the $2! = 2$ orderings of the same pair returns 10. The $k!$ in the denominator is exactly that correction, and it is the whole difference between a permutation count and a combination count.

Two facts fall straight out of the formula. It is symmetric, $\binom{n}{k} = \binom{n}{n-k}$, because choosing which $k$ trials succeed is the same act as choosing which $n - k$ fail, so $\binom{5}{2} = \binom{5}{3} = 10$. And $\binom{n}{0} = \binom{n}{n} = 1$, since there is one way for every trial to fail and one way for every trial to succeed. The distribution built on it is topic 2.10 of Unit 2.