Permutations & Combinations

Both count possibilities, and the difference is whether order matters.

Permutation (order matters, like a race podium or a password):
P(n, r) = n! / (n − r)!

Combination (order does not, like a committee or a lottery ticket):
C(n, r) = n! / (r!(n − r)!)

The extra r! divides out all the rearrangements you do not want to count separately.

From 10 people, choosing 3 for gold/silver/bronze is P(10,3) = 720. Choosing 3 for a committee is C(10,3) = 120 — exactly 6 times fewer, because each committee of 3 can be ordered in 3! = 6 ways.

The test: would swapping two of your picks give a genuinely different outcome?

With n! = n(n−1)…1 and 0! = 1, ordered selections without repetition number P(n,r) = n!/(n−r)! and unordered selections number C(n,r) = n!/(r!(n−r)!).

The relationship C(n,r) = P(n,r)/r! formalises the correction: every unordered set of r elements corresponds to r! distinct orderings, so dividing removes the overcount.

Combinations are the binomial coefficients, which is why Pascal's triangle, the binomial theorem and counting problems are the same mathematics in different notation, and why C(n,r) = C(n, n−r) — choosing which r to include is equivalent to choosing which n−r to exclude.

Repetition changes both formulas: ordered with repetition gives nʳ, and unordered with repetition gives C(n+r−1, r).

In probability, these supply the counts for P(E) = |E|/|S| where outcomes are equally likely, so hypergeometric problems such as card hands are solved by counting favourable combinations over total combinations.