IB Maths AA HLTopic 1 — Number & AlgebraPaper 1 & 2HL only~10 min read
Combinations
A combination is what you use when you’re picking a group and the order of the picks doesn’t matter. Think about choosing three friends to invite to a sleepover — does the order you call them in matter? No. They’re all just “the three you invited”. Compare that to picking the same three friends and giving them gold, silver and bronze medals — now order matters, and that’s a permutation. Same people, but a different question. This note builds on permutations: same starting setup, same factorials, but a small extra division to remove the order. Get comfortable with the nCr formula, learn when to multiply nCr values together for “from each group” problems, and you’ll handle almost every exam question on this topic.
📘 What you need to know
A combination is a selection of r objects from n different ones where order does not matter.
The formula is nCr = n! ÷ [r!(n − r)!]. It’s in the formula booklet.
The big distinction: permutation = order matters; combination = order doesn’t matter.
Useful properties: nC0 = 1, nCn = 1, nC1 = n, and the symmetry nCr = nCn−r.
Connection to permutations: nPr = r! × nCr — a permutation is just a combination followed by arranging the chosen items.
For “from each group” problems (e.g. 2 boys AND 3 girls from a class), multiply the nCr values together.
For “at least” or “at most” problems, split into cases and add the case totals at the end.
What’s a combination?
A combination is a selection — a group you pick out of a bigger pool, where the order of your picks doesn’t change anything. If a teacher chooses Anna, Ben and Cora as the three students who will represent the school at a science fair, the team is the same whether the teacher picked them in the order A → B → C or in the order C → A → B. The trio is the trio.
So the question “how many ways can the teacher choose 3 representatives from 10 candidates?” is a combination question — and the answer involves 10C3, not 10P3.
The big distinction — permutation vs combination
This is where most students lose marks. Both topics use the same setup (“choose r from n“), but the answer differs by a factor of r!. Read the question carefully and ask yourself: does it matter what order I pick them in?
Same 3 people — but two very different questions
Spot the question type from the wording. “Arrange”, “in order”, “1st / 2nd / 3rd”, “President / Vice President” → permutation. “Choose”, “select”, “team”, “committee”, “group” → combination.
If you ever can’t tell whether order matters, ask: “if I swap two of my picks, is it a different outcome?”. Yes → permutation. No → combination. That single question resolves nearly every case.
The nCr formula
Combinations of r from nnCr = n!r!(n − r)! (where 0 ≤ r ≤ n)
✓ in the formula booklet
Same factorial recipe as the permutation formula, plus an extra r! in the denominator. That extra division is the whole reason the answer is smaller — it strips out all the orderings of the chosen group.
Quick worked-through example: how many ways can you choose 4 people for a committee from 12? Substitute n = 12 and r = 4:
Notice the cancellation trick: write 12! as 12·11·10·9·8! and the 8! divides out the bottom one. So you only multiply 12 × 11 × 10 × 9 on top, and 4! = 24 on the bottom. Don’t ever multiply 12! out fully — that’s 479 million pointless digits.
Four properties that save you time
Just like with the binomial coefficient, there are a few small facts about nCr that turn up in exams again and again. Worth committing to memory.
nC0 = 1
There’s exactly one way to choose nothing — by choosing nothing.
nCn = 1
There’s exactly one way to choose everything — by taking the whole pool.
nC1 = n
Choosing one out of n just gives n options. No cleverness needed.
nCr = nCn−r
Symmetry. Choosing r to take is the same as choosing n − r to leave behind.
Symmetry shortcut:20C17 looks awful, but it equals 20C3 = (20·19·18)/(3·2·1) = 1140. Always check the smaller of r and n − r before computing.
How nPr and nCr are connected
Compare the two formulas:
nPr = n!(n − r)! vs nCr = n!r!(n − r)!
The only difference is that extra r! in the denominator of nCr. Rearrange and you get a really clean relationship:
Permutations vs combinationsnPr = r! × nCr
🤔 Why does this make sense?
Picking r people and arranging them in order is a two-step process: first choose the group (nCr), then arrange that group in a row (r! ways). Multiply the two and you get the total ordered selections — which is exactly nPr. So a permutation is just “a combination that’s been arranged”.
Multiplying nCr values — “from each group” problems
Lots of exam questions ask you to choose a few items from one group AND a few from another. Whenever you see “AND”, reach for the multiplication rule: count each group separately with nCr, then multiply.
🧭 Recipe — picking from multiple groups
Identify each group the question gives you (e.g. boys, girls, fantasy books, classics).
Work out how many you need from each using the question’s wording.
Compute the nCr for each group separately.
Multiply them all together — that’s the AND rule from counting principles.
Example sketch: a class has 8 boys and 6 girls. You want a team of 2 boys AND 3 girls. Number of ways = 8C2 × 6C3 = 28 × 20 = 560.
“At least” and “at most” — using cases
When a question says “at least 2 of these” or “no more than 3 of those”, you usually have to split into cases. Each case is one possible split (e.g. exactly 2, exactly 3, exactly 4), and you add the case totals at the end.
🧭 Recipe — “at least” / “at most”
List every valid case that satisfies the condition. For “at least 2 red”, that’s 2 red, 3 red, 4 red, … up to whatever’s possible.
For each case, multiply nCr values using the from-each-group recipe.
Add the case totals.
Sanity check: the case r-values should add up to the total number of items being chosen.
A handy shortcut for “at least” problems: sometimes it’s faster to use the complement. “At least 1 X” is much quicker as (total selections) − (zero X selections). It works the same way as the “must not be together” trick from the permutations note — count the easy opposite, subtract.
Worked examples
WE 1
Choosing students for a school trip
A teacher needs to choose 5 students from a group of 12 to go on a school trip. In how many different ways can the teacher make this selection?
Step 1: Order doesn’t matter — it’s a combinationuse 12C5 with n = 12, r = 5Step 2: Apply the formula and cancel the bigger factorial12C5 = 12! / (5! · 7!)= (12 · 11 · 10 · 9 · 8) / 5!top: 12 · 11 = 132, 132 · 10 = 13201320 · 9 = 11880, 11880 · 8 = 95040Step 3: Divide by 5! = 12095040 / 120 = 792792 different selectionsa group of 5 is a group of 5 — swapping two students gives the same group
WE 2
Permutation vs combination — same setup, different question
A club has 10 members. Find the number of ways to choose 3 of them as:
(a) a President, Vice-President, and Secretary,
(b) a 3-person planning committee.
Part (a): order matters (different roles) → permutation10P3 = 10! / 7! = 10 · 9 · 8 = 720(a) 720 waysPart (b): order doesn’t matter (same committee) → combination10C3 = (10 · 9 · 8) / 3! = 720 / 6 = 120(b) 120 wayspart (a) is exactly 3! = 6 times bigger than part (b) — that’s nPr = r! × nCr in action
WE 3
Pizza — picking from each group
A pizza shop offers 6 different toppings and 4 different cheeses. A customer wants to design a pizza with 3 toppings and 2 cheeses. How many different pizzas can be made?
Step 1: Spot the AND — 3 toppings AND 2 cheesesmultiply nCr valuesStep 2: Compute each nCrtoppings: 6C3 = (6 · 5 · 4)/(3 · 2 · 1) = 120/6 = 20cheeses: 4C2 = (4 · 3)/(2 · 1) = 6Step 3: Multiplytotal = 20 × 6 = 120120 different pizzasorder doesn’t matter for pizza toppings — pepperoni-then-mushroom is the same pizza as mushroom-then-pepperoni
WE 4
“At least” — splitting into cases
A jar contains 5 red marbles and 7 blue marbles. Four marbles are picked at random. In how many ways can the selection contain at least 2 red marbles?
Step 1: List the valid cases — “at least 2 red” of 4 picks means 2, 3, or 4 redCase A: 2 red + 2 blueCase B: 3 red + 1 blueCase C: 4 red + 0 blueStep 2: Compute each case using the AND ruleCase A: 5C2 × 7C2 = 10 × 21 = 210Case B: 5C3 × 7C1 = 10 × 7 = 70Case C: 5C4 × 7C0 = 5 × 1 = 5Step 3: Add the case totals (OR rule)total = 210 + 70 + 5 = 285285 selectionsalways check: 2+2 = 4, 3+1 = 4, 4+0 = 4 ✓ — every case picks exactly 4 marbles
WE 5
The handshake problem
Twelve people are at a meeting. Every person shakes hands with every other person exactly once. How many handshakes take place in total?
Step 1: Each handshake is a pair of people — order doesn’t matter“A shakes with B” is the same handshake as “B shakes with A”so it’s a combination: choose 2 from 12Step 2: Compute 12C212C2 = (12 · 11) / 2! = 132 / 2 = 6666 handshakesclassic application — for n people in a room, the total handshake count is nC2
💡 Top tips
Read the question for “order matters” clues. Words like “arrange”, “rank”, “in order” → permutation. “Choose”, “select”, “team”, “committee” → combination.
Use symmetry to halve the work.20C18 = 20C2 — much easier. Always go for the smaller of r and n − r.
Cancel the bigger factorial first.12C4 = (12·11·10·9) / 4! — never multiply 12! out fully.
For “from each group” problems, multiply. Compute each nCr separately, then multiply them — that’s the AND rule.
For “at least” or “at most”, split into cases. Each case is a specific split; compute each, then add. The case r-values should always sum to the total picks.
Use the complement trick when it’s faster. “At least 1 X” = (total) − (zero X). Sometimes much quicker than listing cases.
Use your GDC. The nCr button (or “C”) is faster than computing factorials by hand on Paper 2.
⚠ Common mistakes
Using nPr when order doesn’t matter. “Pick 3 students for a team” is a combination, not a permutation. Picking a team and arranging that team in a row are different questions.
Using nCr when order does matter. Awarding gold/silver/bronze is a permutation — the order genuinely changes the outcome.
Forgetting to multiply across groups. “2 boys and 3 girls” needs both nCr values, multiplied together — not added.
Adding within a single AND scenario. “Choose a captain AND a vice-captain” needs multiplication, not addition.
Skipping a case in “at least” problems. If “at least 2 red” out of 4 picks, you need exactly 2, 3, AND 4 cases — missing any one drops marks.
Not checking that case r-values sum correctly. If you’re choosing 4 marbles and a case says “5 red + 0 blue”, you’ve made a mistake — the picks per case must total 4.
Forgetting that 0! = 1. If you treat 0! as 0, nC0 and nCn blow up. They should both equal 1.
Combinations look harder than permutations because the questions are wordier and the cases multiply quickly — but the underlying machinery is exactly the same. Read carefully, ask “does order matter?”, and remember that “AND across groups” means multiply while “at least” means split into cases. With practice, the structure of these questions becomes obvious within seconds of reading them. That wraps up the Permutations & Combinations section — next up is Complex Numbers, a brand-new topic where the rules of arithmetic suddenly stretch beyond the real number line.
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