Combinations: Choosing When Order Does Not Matter

What is a combination?

A combination is an unordered selection. If the same group of people in a different order counts as the same outcome, you are counting combinations.

The symbol $C(n, r)$ — also written $\binom{n}{r}$ or $_nC_r$ — is read "$n$ choose $r$."

The formula

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

The last form shows what's really going on: combinations are permutations with the redundant orderings divided out.

Step-by-step solution

We want to pick a committee of 3 from 8 students. A committee is a set, so {Ava, Ben, Cal} is the same committee as {Ben, Cal, Ava}.

Step 1 — Count as if order mattered: $P(8, 3) = 8 \cdot 7 \cdot 6 = 336$

Step 2 — How many orderings of each committee? Any 3 people can be arranged in $3! = 6$ ways. So every committee was counted 6 times above.

Step 3 — Divide by $r!$ to remove the overcounting: $C(8, 3) = \dfrac{336}{6} = \boxed{56}$

Step 4 — Verify with the factorial formula: $C(8,3) = \dfrac{8!}{3! \, 5!} = \dfrac{40320}{6 \cdot 120} = \dfrac{40320}{720} = 56 \checkmark$

Why we divide by $r!$

Count the permutations {A, B, C}, {A, C, B}, {B, A, C}, {B, C, A}, {C, A, B}, {C, B, A} — that's $3!=6$ different permutations but only 1 committee. Dividing by $3!$ collapses each group of 6 down to 1.

Symmetry: $C(n, r) = C(n, n-r)$

Choosing the 3 committee members is the same as choosing the 5 people to leave out. Indeed $C(8, 3) = C(8, 5) = 56$. This shortcut helps when $r$ is close to $n$.

Common mistakes

• Using permutations when order does not matter. Committees, teams, hands of cards, subsets — all are combinations.
• Forgetting to divide by $r!$. This is the #1 error. Always ask: "Does ABC count the same as CBA?" If yes, divide by $r!$.
• Miscomputing factorials. $0! = 1$ by convention, and $C(n, 0) = C(n, n) = 1$ (there is exactly one way to choose nothing or everything).

Try it in the visualization

Slide $n$ and $r$ — the grid highlights one combination at a time, cycling through all $C(n, r)$ of them. A counter at the bottom shows both $P(n,r)$ and $C(n,r)$ so you can see the factor-of-$r!$ difference.