Only who was chosen, not the order — divide the line-ups back out.
In the last lesson you lined people up, and the order mattered: A–B was a different photo than B–A. But most of life's "how many ways?" questions don't care about order at all. A handshake is the same handshake whoever reaches out first. A committee of three is the same committee no matter who you name first. A hand of cards is the same hand however it's dealt. When once the group is chosen you are done — when shuffling the order changes nothing — you are counting combinations, not permutations. And the trick is delightfully simple: count the ordered line-ups you already know how to count, then divide the order back out.
Picture three friends — Ann, Ben, Cara — and you must pick two of them to carry a banner. List the choices: {Ann, Ben}, {Ann, Cara}, {Ben, Cara}. That's it — three pairs. Notice what we did not write: {Ben, Ann} is not a fourth pair, because it is the very same two people. Once the group is chosen, who was named first makes no difference.
That is the whole idea. A combination is a selection without order — you keep who was chosen and throw away the sequence. We write it C(n, m): the number of ways to choose m things out of n when order does not matter. (On a calculator it's the ₙCᵣ button, and out loud people say "n choose m." Some books stack the two numbers in big parentheses.) Compare that with a permutation P(n, m), where the line-up did matter.
A combination is a choice with the order thrown away. C(n, m) counts the ways to pick m from n when shuffling the chosen ones makes no new choice. Handshakes, committees, card hands, lottery tickets, subsets — all combinations.
Here is the elegant part. We already know how to count ordered picks: P(n, m) line-ups. And every single committee of m people shows up among those line-ups exactly m! times — once for each way to arrange the very same m people. So the ordered count is m! times too big. To get the unordered count, divide by m!:
C(n, m) = P(n, m)m! = n!m! (n − m)!
Try it on our three friends. Ordered: P(3, 2) = 3 × 2 = 6 line-ups. But each pair was counted 2! = 2 times (AB and BA), so C(3, 2) = 6 / 2 = 3 — exactly the three pairs we listed. And a bigger one: choosing 3 of 5, the line-ups are P(5, 3) = 5 × 4 × 3 = 60, each committee counted 3! = 6 times, so C(5, 3) = 60 / 6 = 10.
If you used P(n, m) for a committee, you counted every committee m! times. A committee has no first chair, no second chair — there is no order to keep. Always divide by m! to undo the line-ups you don't want.
A famous one: a poker hand is 5 cards chosen from 52, and the order you're dealt them is irrelevant. So the number of distinct hands is C(52, 5) = 2,598,960 — more than two and a half million, all from one little division.
To pick m people to include is the very same act as picking the n − m people to leave out — every choice of one decides the other. So the two counts must be equal:
C(n, m) = C(n, n − m)
For instance C(7, 2) = 21 and C(7, 5) = 21 — same number, because choosing 2 to keep is choosing 5 to drop. This is also a handy shortcut: to find C(20, 18), just compute the easy twin C(20, 2) = 190.
Now suppose one more person — call them the newcomer — joins a group of n, and you still want a committee of m from the n + 1 total. Split into two cases that can't both happen:
Add the disjoint cases (32.1's Addition Principle), and you get one of the loveliest facts in mathematics:
C(n + 1, m) = C(n, m) + C(n, m − 1)
Each "choose" number is the sum of the two above it. Check it: C(6, 2) = 15, and C(5, 2) + C(5, 1) = 10 + 5 = 15. That "add the two above" pattern is exactly Pascal's triangle — the spine of the next lesson, 32.4 · The Binomial Theorem.
Beginners agonize over which formula to use. There is a single test that settles it every time. Ask yourself:
"If I shuffle the order of what I chose, is it still the same outcome?"
"Choose 3 students for a committee" — shuffle them, same committee → C(n, 3). "Choose 3 students for president, secretary, treasurer" — shuffle them, different jobs, different outcome → P(n, 3). "Deal a 5-card hand" → combination. "Set a 3-digit code from distinct digits" → permutation. Same people, same numbers — the only thing that changes is whether the sequence carries meaning.
Treating {A, B} and {B, A} as two different committees is the permutation mistake wearing a combination's clothes. If shuffling makes nothing new, you must divide the order out.
Real problems add conditions: "this person must be on the team," "use at most one senior." Two reliable tactics handle nearly all of them.
Say a club of 8 must form a committee of 3, and Ann must be on it. Put Ann on first — that seat is decided. Now you only choose the remaining 2 from the other 7: C(7, 2) = 21 committees. ("Must exclude" is the same idea: drop that person and choose from the rest.)
A team of 4 is chosen from 2 seniors and 6 juniors, with the rule at most one senior. There are two honest ways to count it.
Direct, by cases (add the disjoint possibilities):
Total = 15 + 40 = 55.
Total minus bad (count everything, subtract what breaks the rule):
Total = 70 − 15 = 55 — the same answer, reached from the other side. Pick whichever path has fewer cases.
When the restriction is "at least one," the total − bad route is usually shortest: count all selections, then subtract the ones with none of the required type. One subtraction beats a stack of cases.
Last idea, and it's where people slip. Suppose you split 4 people into two groups of 2. You might reach for C(4, 2) · C(2, 2) = 6 × 1 = 6. But the two groups are both size 2 and unlabeled — splitting into {A,B} & {C,D} is the same split as {C,D} & {A,B}. You counted each split twice. There are 2 equal piles, so divide by 2!:
C(4, 2) · C(2, 2)2! = 6 · 12 = 3
The rule: when you form several groups that are equal in size and interchangeable, divide by the factorial of the number of equal piles. With three equal piles of 2 from 6 people, it's C(6,2)·C(4,2)·C(2,2) / 3! = (15 × 6 × 1)/6 = 15 splits.
If the piles are distinguishable — handed to three named people, or labeled Room A / Room B / Room C — then do not divide; the order among piles now matters. Distributing 6 people, 2 to each of three named rooms, is C(6,2)·C(4,2)·C(2,2) = 90. Only identical, unlabeled piles get the extra division.
| Situation | What to do | Why |
|---|---|---|
| Choose m of n, order doesn't matter | C(n, m) = P(n, m)/m! = n!/(m!(n−m)!) | Each committee was counted m! times as line-ups — divide it out |
| Swap "in" for "out" | C(n, m) = C(n, n − m) | Choosing who's in decides who's out |
| A newcomer is in or out | C(n+1, m) = C(n, m) + C(n, m−1) | Pascal's "add the two above" → 32.4 |
| Permutation or combination? | Shuffle the order — same choice? → C. New choice? → P | The one test that always works |
| A required member | Seat them first, choose the rest | One seat fixed → C(n−1, m−1) |
| "At most / at least k" | Direct cases (add), or total − bad | Take the route with fewer cases |
| Equal, interchangeable piles | Divide by (# equal piles)! | Identical piles otherwise counted twice |
A class of 10 students sends a delegation of 4 to a conference (no special roles). How many delegations are possible?
There are 9 people at a reunion and everyone shakes hands once with everyone else. How many handshakes happen?
Without computing both, explain why C(20, 17) equals C(20, 3), then give the value.
A committee of 3 is chosen from 8 people, but Maria must be included. How many committees?
From 2 seniors and 6 juniors, a team of 4 is formed with at least one senior. How many teams? (Use total − bad.)
Six players are split into three teams of two for a doubles tournament (the teams are not named or ranked). How many ways?
Six questions to lock it in. Tap the answer you think is right.
This lesson builds combinations as permutations with the order divided out — C(n, m) = P(n, m)/m! = n!/(m!(n−m)!) — and develops the symmetry C(n, m) = C(n, n − m) and Pascal's recurrence C(n+1, m) = C(n, m) + C(n, m−1), the bridge to Stage 32.4's Binomial Theorem. It directly supports CCSS HSS-CP.B.9 — use permutations and combinations to compute probabilities of compound events and solve problems — and the single "does shuffling change the outcome?" test gives students a dependable way to choose between the two. A good follow-up conversation: ask your learner to invent one situation that needs C and one that needs P using the very same people, and to say out loud which one shuffling would change.