A Level Maths 9709 · Paper 5 Probability & Statistics 1 · 5.2 · ~4 min read
Permutations and combinations
Counting arrangements with factorials, repeated items, items kept together or apart, and selections with nCr, including "at least" conditions and probabilities found by counting.
FactorialsArrangementsRepeated lettersTogether & apartPermutations ⁿPᵣCombinations ⁿCᵣRestrictions"At least"
RSby Riyaz Sir
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Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end
In 30 seconds
- Arrangements of n different items: n!. With repeats, divide by the factorial of each repeat count.
- Order matters → permutations ⁿPᵣ; order doesn't matter → combinations ⁿCᵣ.
- Together: glue the group into one block, arrange the blocks, then arrange inside the block. Apart: arrange the others first, then place the separated items in the gaps.
- "At least" questions: add the separate cases, or subtract the unwanted cases from the total.
By the end you can
- Count arrangements of objects in a line, including repeated objects.
- Solve problems with restrictions: items together, apart, or in fixed positions.
- Count selections using combinations, including with restrictions.
- Find probabilities by counting arrangements or selections.
1Arrangements
Factorials and repeats
n different items: n! ; with repeats: n!/(p! q! …) ; ⁿPᵣ = n!/(n − r)!
ⁿPᵣ is r items chosen from n, in orderWorked example
How many arrangements are there of the letters of BANANA?
- 6 letters, with A three times and N twice M1
- 6!/(3! 2!) = 60 A1
Worked example
The 7 letters of ORANGES are arranged. Find the number of arrangements in which (a) the vowels are together and (b) no two vowels are next to each other.
- (a) block (OAE) + R, N, G, S = 5 items: 5! × 3! = 720 M1 A1
- (b) arrange the consonants: 4! = 24; 5 gaps, place 3 vowels in order: ⁵P₃ = 60 M1
- 24 × 60 = 1440 A1
- Fixed positions: deal with the restricted positions first. "Begins and ends with S" fixes those two places.
Common mistake"Not together" = total − together only works for two items. For three vowels, "no two together" is not "not all three together". The 2024 Paper 5 report noted this confusion.
2Selections
Combinations
ⁿCᵣ = n!/[r!(n − r)!]
the number of ways to choose r from n when order doesn't matterWorked example
A committee of 5 is chosen from 6 men and 4 women. How many committees contain at least 2 women?
- total ¹⁰C₅ = 252 M1
- no women: ⁶C₅ = 6; one woman: 4 × ⁶C₄ = 60 M1
- 252 − 6 − 60 = 186 A1
Worked example
In the committee question, two particular people refuse to serve together. How many committees are possible (no other condition)?
- committees with both: ⁸C₃ = 56 M1
- 252 − 56 = 196 A1
- Probability by counting: P = (number of favourable selections)/(total selections). For a random committee, P(at least 2 women) = 186/252 = 0.738.
Common mistakeMultiplying cases that should be added. "2 women or 3 women" are separate cases: add them. Multiply only for choices made together.
Linking questionCombinations give the binomial coefficients in Discrete random variables and the expansion in Series; probabilities by counting are in Probability.
Test yourself — cover the answers first
Quick check
- Evaluate ⁷C₃.
- How many arrangements of 5 different books on a shelf?
- How many arrangements of the letters of LEVEL?
- Choose a captain and vice-captain from 11 players.
- Choose 3 people from 8 (order doesn't matter).
Exam-style
- How many arrangements of STATISTICS are there?
- 4 boys and 3 girls stand in a line. How many ways if the girls are all together?
- A team of 4 is chosen from 5 Year 12 and 6 Year 13 students. Find the probability it has exactly 2 from each year.
Show answers
- 35
- 120
- 5!/(2! 2!) = 30
- ¹¹P₂ = 110
- ⁸C₃ = 56
- 10!/(3! 3! 2!) = 50 400
- 5! × 3! = 720
- ⁵C₂ × ⁶C₂/¹¹C₄ = 150/330 = 0.455
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