Combinations

A combination is an unordered selection — a handful, a committee, a lottery ticket — where rearranging the chosen items changes nothing. Combinations are permutations with the ordering divided out, and they're everywhere in probability.

Build the intuition

Selection without order

Choosing 2 friends from 5 to invite: {Ana, Ben} is the same invitation as {Ben, Ana}. Order is irrelevant, so we count selections, not arrangements. The notation is 'n choose k', written C(n, k) or the binomial coefficient. A committee, a hand of cards, a subset — all combinations.

\binom{n}{k} = \frac{n!}{k!\,(n-k)!}

Divide out the orderings

Start from permutations: P(n, k) counts ordered selections. But each unordered group of k items was counted k! times — once for every ordering of those same k. So divide by k! to collapse them into one: C(n, k) = P(n, k) / k!. That single division is the entire difference between permutations and combinations.

\binom{n}{k} = \frac{P(n,k)}{k!}

The deciding question

Always ask first: would rearranging the chosen items create a different outcome? Podium (gold vs silver) — yes, use permutations. Pizza toppings, committee members, lottery numbers — no, use combinations. Getting this one question right is the difference between a correct count and one that's off by a factor of k!. When unsure, picture two specific orderings and ask if they're 'the same'.

See it move

InteractiveThe counting lab

Stages2

Choices each3

2 stages, each with 3 choices → 3^2 = 9 outcomes. The fundamental counting principle: independent choices multiply.

Switch to 'Order doesn't matter': the permutation count appears, then gets divided by k! to merge the re-orderings of each group into one. That division is the whole idea.

A worked example

Count the handshakes

8 people at a party each shake hands once with every other. How many handshakes?
A handshake is an unordered pair — choose 2 from 8, order irrelevant:
$\binom{8}{2} = \frac{8 \times 7}{2} = 28$
We divided the 56 ordered pairs by 2! = 2, since A-shakes-B is the same handshake as B-shakes-A.
28 handshakes. Use permutations here and you'd double-count every one.

Out in the world

Lottery odds, exactly

Pick 6 numbers from 49, order irrelevant: C(49, 6) = 13,983,816 possible tickets — so one ticket has a 1-in-14-million chance. Lotteries are pure combinations, and the formula is, in effect, the consumer-warning label on the ticket.

Common confusion, cleared

“C(n, k) is bigger than P(n, k).”

It's smaller — by a factor of k!. Combinations merge the orderings that permutations count separately, so there are always fewer (or equal) combinations than permutations.

“Choosing k to keep is different from choosing k to leave out.”

They're equal: C(n, k) = C(n, n−k). Picking 6 lottery numbers to play is the same count as picking 43 to skip. The symmetry often simplifies the arithmetic.

Check yourself

PracticeQuick check

A pizza lets you choose 3 toppings from 10. How many topping combinations?
Which situation needs combinations (not permutations)?

Recap

A combination is an unordered selection; rearranging changes nothing.
C(n, k) = P(n, k) / k! — divide out the k! orderings.
Ask 'does order matter?' first; C(n, k) = C(n, n−k).

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