PERMUTATIONS & COMBINATIONS

The two formulas

• Permutations (order matters): the number of ways to arrange $r$ items chosen from $n$ distinct items is $P(n, r) = \frac{n!}{(n-r)!}.$
• Combinations (order doesn't matter): the number of ways to choose $r$ items from $n$ distinct items is $\binom{n}{r} = \frac{n!}{r!\,(n-r)!}.$

The relationship: $\binom{n}{r} = \dfrac{P(n,r)}{r!}$ — divide out the orderings you're not distinguishing.

When to use which

• "How many ways to arrange / line up / order?" → permutation.
• "How many ways to choose / pick a team / a subset?" → combination.
• "How many distinct seating arrangements around a round table?" → circular permutation, which is $(n-1)!$ because rotations are equivalent.

Worked example

A 4-person committee is chosen from a group of 10. How many different committees are possible?

Order doesn't matter (committees are unordered), so the answer is $\binom{10}{4} = \frac{10!}{4!\,6!} = 210.$

Common trap

If the problem says "the committee has a chair, a treasurer, and two members," you are picking roles — that's an ordered situation in disguise. Use permutations (or pick the chair, then the treasurer, then a combination for the remaining two).