Permutations & Combinations

Factorials

$n! = n \cdot (n-1) \cdot (n-2) \cdots 2 \cdot 1$

By convention $0! = 1$. Examples: $5! = 120$, $4! = 24$.

A factorial counts the ways to arrange $n$ distinct items in a row.

Permutations — when order matters

A permutation counts ordered arrangements. How many ways can you arrange $r$ items chosen from $n$?

$P(n, r) = \frac{n!}{(n-r)!}$

Example. A class of 8 students picks a president, vice president, and treasurer (three different roles). How many ways?

$P(8, 3) = \frac{8!}{5!} = 8 \cdot 7 \cdot 6 = 336$

Combinations — when order doesn't matter

A combination counts unordered selections. How many ways to choose $r$ items from $n$, ignoring order?

$C(n, r) = \binom{n}{r} = \frac{n!}{r!\,(n - r)!}$

The extra $r!$ in the denominator divides out the duplicates created by reordering the chosen group.

Example. Same class of 8 picks a 3-person planning committee (no distinct roles).

$\binom{8}{3} = \frac{8!}{3! \cdot 5!} = \frac{8 \cdot 7 \cdot 6}{6} = 56$

That's $336 / 6 = 56$ — exactly the permutation count divided by $3! = 6$, the number of ways to reorder 3 chosen people.

Quick test: order or no order?

• "How many license plates?" → spots matter → permutation.
• "How many 5-card poker hands?" → cards have no order → combination.
• "How many ways to seat 5 people in 5 chairs?" → permutation ($5! = 120$).
• "How many 3-topping pizzas from 10 toppings?" → combination.

Useful identities

$\binom{n}{0} = \binom{n}{n} = 1$ $\binom{n}{r} = \binom{n}{n-r}$ $\binom{n}{1} = n$

Key takeaways

• Permutations count ordered arrangements; combinations count unordered selections.
• Combinations divide out the $r!$ reorderings of the chosen group.
• Decide by asking: "does swapping two items give a different outcome?"