Combinations
Choosing 2 students from 5 to go on a trip: choosing Asha and Bela is the same as Bela and Asha. Only who is chosen matters, not the order.
A combination is a selection of objects where order does not matter. The number of ways to choose r from n distinct objects is ⁿCᵣ = n!/(r!(n − r)!) = ⁿPᵣ/r!.
Formula: ⁿCᵣ = n! / (r! (n − r)!)
Rule 1: Compute ⁿCᵣ — Write r factors going down from n, divide by r!.
- Easy: ⁶C₂ = (6 × 5)/(2 × 1) = 15.
- Medium: ¹⁰C₃ = (10 × 9 × 8)/(3 × 2 × 1) = 120.
- Exam-type: ⁸C₈ = 1 and ⁸C₀ = 1.
Rule 2: Choose a group — Use ⁿCᵣ when the order of the chosen group does not matter.
- Easy: A team of 3 from 7 players: ⁷C₃ = 35.
- Medium: Choose 2 books from 5: ⁵C₂ = 10.
- Exam-type: Handshakes among 6 people: ⁶C₂ = 15.
ⁿC₁ = n, ⁿC₀ = 1, ⁿCₙ = 1.
Using ⁿPᵣ when only the group matters.
C is for Committee: order does not matter.
Spot the difference
| ⁿPᵣ | ⁿCᵣ |
|---|---|
| Order matters | Order does not matter |
| Bigger value | Smaller by r! |
Practice questions
- 1Find ⁵C₂.
- 2Find ⁸C₃.
- 3Find ⁹C₉.
- 4In how many ways can 4 students be chosen from 9?
- 5How many diagonals does a hexagon have? (⁶C₂ − 6)
- 6If ⁿC₂ = 45, find n.
- 7How many triangles can be drawn using 7 points, no three on a line?
- 8If ⁿC₃ = ⁿC₂, find n.
Show answers (try them yourself first)
- 110
- 256
- 31
- 4126
- 59
- 6n(n − 1) = 90, so n = 10
- 735
- 83 + 2 = n, so n = 5
Properties of ⁿCᵣ
Choosing 3 students to go is the same as choosing the 3 who stay home (when there are 6 in all).
Key results: ⁿCᵣ = ⁿCₙ₋ᵣ; ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ; if ⁿCₓ = ⁿCᵧ then x = y or x + y = n.
Formula: ⁿCᵣ = ⁿCₙ₋ᵣ
Rule 1: Use symmetry — Pick the smaller of r and n − r.
- Easy: ¹⁰C₈ = ¹⁰C₂ = 45.
- Medium: ¹⁵C₁₃ = ¹⁵C₂ = 105.
- Exam-type: ⁹C₆ = ⁹C₃ = 84.
Rule 2: Pascal rule — Adjacent entries add up.
- Easy: ⁵C₂ + ⁵C₃ = ⁶C₃ = 20.
- Medium: ⁷C₃ + ⁷C₄ = ⁸C₄ = 70.
- Exam-type: ⁴C₀ + ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄ = 2⁴ = 16.
The sum of all ⁿCᵣ for r = 0 to n is 2ⁿ.
Writing ⁿCᵣ = ⁿC(r − n).
Choose or leave: both counts agree.
Practice questions
- 1Find ¹²C₁₀.
- 2Find ⁶C₂ + ⁶C₃.
- 3If ⁿC₄ = ⁿC₆, find n.
- 4If ⁿC₈ = ⁿC₂, find ⁿC₂.
- 5Find ⁴C₀ + ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄.
- 6Find ¹⁰C₃ + ¹⁰C₄ using Pascal rule.
- 7If ²ⁿC₃ : ⁿC₃ = 11 : 1, find n.
- 8Prove that ⁿCᵣ × r!(n − r)! = n!.
Show answers (try them yourself first)
- 166
- 235 = ⁷C₃
- 3n = 10
- 4n = 10, ⁿC₂ = 45
- 516
- 6¹¹C₄ = 330
- 7(2n)(2n−1)(2n−2)/(n(n−1)(n−2)) = 11 gives 4(2n − 1)/(n − 2) = 11, so n = 6
- 8Rearrange the definition ⁿCᵣ = n!/(r!(n − r)!)
Selection problems
A cricket team of 11 from 15 players, but the captain must be in it: one place is fixed, so we choose the other 10 from 14.
Fixed members: include them and choose the rest. Excluded members: remove them from n. "At least one": total − none.
Rule 1: Include or exclude — Adjust n and r.
- Easy: Committee of 4 from 8 with a particular person included: ⁷C₃ = 35.
- Medium: Committee of 4 from 8 with a particular person excluded: ⁷C₄ = 35.
- Exam-type: Choose 3 from 5 boys and 4 girls: ⁹C₃ = 84.
Rule 2: Mixed groups and "at least" — Multiply for AND, subtract for at least one.
- Easy: 2 boys and 2 girls from 5 boys and 4 girls: ⁵C₂ × ⁴C₂ = 60.
- Medium: At least one girl in a group of 3 from 5 boys and 4 girls: 84 − ⁵C₃ = 74.
- Exam-type: Choose 3 cards from 13 hearts: ¹³C₃ = 286.
Count groups as AND (multiply) when picking from different sets.
Counting groups twice when "at least" cases overlap.
At least → total minus none.
Practice questions
- 1A group of 3 is chosen from 6 students. How many groups?
- 2A group of 3 from 6 students with a given student always in.
- 32 boys and 1 girl from 4 boys and 3 girls.
- 4A group of 4 from 5 men and 4 women with exactly 2 women.
- 5A group of 4 from 5 men and 4 women with at least one woman.
- 6A group of 5 from 6 boys and 4 girls with at most 2 girls.
- 7How many ways to choose 5 cards from 52 cards that are all spades?
- 8A box has 5 red and 4 white balls. In how many ways can 3 balls be drawn so that all 3 are the same colour?
Show answers (try them yourself first)
- 120
- 210
- 318
- 460
- 5126 − 5 = 121
- 66 + 60 + 120 = 186
- 71287
- 810 + 4 = 14
Permutation or combination?
Choosing 3 players is a combination; choosing who will bat first, second and third is a permutation. We often do both: first choose, then arrange.
Ask: does the order matter? If yes use ⁿPᵣ, if no use ⁿCᵣ. When both happen, choose first, then arrange: ⁿCᵣ × r! = ⁿPᵣ.
Formula: ⁿPᵣ = ⁿCᵣ × r!
Rule 1: Decide — Look for words like arrange, line up, order (P) and choose, select, committee (C).
- Easy: Choose 3 captains from 8: ⁸C₃.
- Medium: Choose a captain, a vice-captain and a coach from 8: ⁸P₃.
- Exam-type: Choose a committee of 3 and arrange them in a row: ⁸C₃ × 3!.
Rule 2: Choose then arrange — Multiply the two counts.
- Easy: Choose 3 letters from 6 and arrange them: 20 × 6 = 120.
- Medium: Select 2 boys and 2 girls and seat them in a row: ⁴C₂ × ⁴C₂ × 4! = 864.
- Exam-type: Words of 3 letters with 2 vowels and 1 consonant from {A, E, I, B, C}: ³C₂ × ²C₁ × 3! = 36.
ⁿPᵣ is always r! times ⁿCᵣ.
Using ⁿPᵣ for committees.
Pick first, Position later.
Practice questions
- 1Choose 2 prize winners (same prize) from 7.
- 2Give 1st and 2nd prizes to 2 of 7 students.
- 3Choose 3 of 6 people and seat them in a row.
- 4How many 3-digit numbers can be formed using 3 different digits from 1 to 6?
- 5Choose 2 vowels from 5 and 2 consonants from 4 and arrange the 4 letters.
- 6A committee of 3 men and 2 women from 6 men and 5 women, and then a chairperson chosen from them.
- 7How many 4-letter words (distinct letters) can be formed from MONDAY with exactly 2 vowels?
- 8Show that ¹⁰P₃ = 6 × ¹⁰C₃.
Show answers (try them yourself first)
- 121
- 242
- 3120
- 4120
- 510 × 6 × 4! = 1440
- 6200 × 5 = 1000
- 7Vowels are O, A; consonants M, N, D, Y: ²C₂ × ⁴C₂ × 4! = 144
- 8720 = 6 × 120 = 720