Permutation vs Combination

Permutations and combinations look almost identical until you ask one question: does order matter? Get that wrong and your probability answer will be off by a factor of $r!$ or more. Here's the clean distinction with worked examples.

The core question: does order matter?

• Yes, order matters → permutation. Picking 1st / 2nd / 3rd place from 10 runners.
• No, order doesn't matter → combination. Picking a 5-person committee from 20 people.

Same 10 candidates can give different answers depending on whether the roles are distinct.

The formulas

For $n$ items, choose $r$:

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

Notice combination is permutation divided by $r!$ — that $r!$ removes the orderings of the chosen items, since combinations don't care about order.

Worked examples

Permutation: race podium

Ten runners, three medal positions (gold, silver, bronze). Order matters — gold ≠ silver.

$P(10, 3) = \frac{10!}{7!} = 10 \times 9 \times 8 = 720.$

Combination: lottery numbers

Pick 6 numbers from 49 — the order on your ticket doesn't matter.

$C(49, 6) = \binom{49}{6} = \frac{49!}{6! \cdot 43!} = 13{,}983{,}816.$

Same numbers, different answer

Pick 3 letters from {A, B, C, D}.

• As permutation (3-letter passwords): $P(4, 3) = 24$. ABC, ACB, BAC, ... all distinct.
• As combination (just choosing 3 letters): $C(4, 3) = 4$. {A,B,C}, {A,B,D}, {A,C,D}, {B,C,D}.

The factor of $3! = 6$ between them is exactly the $r!$ in the formula.

Decision shortcut

When in doubt, ask: "If I swap two of my chosen items, is the result different?"

• Yes → permutation
• No → combination

Choosing a captain and vice-captain → swapping changes who's captain → permutation.
Choosing 2 people for a duo → swapping is the same duo → combination.

Common mistakes

• Mixing the two when probability is involved. The denominator (total outcomes) and numerator (favourable outcomes) must use the same counting method.
• Forgetting the $r!$ divisor. If you compute permutations when you wanted combinations, you'll overcount by $r!$.
• Distinguishable vs indistinguishable items. If some items are identical (e.g. 5 red balls and 3 blue), neither plain formula applies — you need the multinomial coefficient $\frac{n!}{n_1! n_2! \cdots}$.

Try it yourself

Use our Probability Calculator to compute permutations, combinations, and apply them to real probability problems with the AI walking you through every step.