A committee, a team or a hand of cards is a selection: the same people in a different order is the same choice. Combinations count selections, and most Paper 5 questions add a condition that means splitting into cases.
The number of ways of choosing \(r\) objects from \(n\) different objects, when order does not matter, is \({}^{n}C_{r} = \dbinom{n}{r} = \dfrac{n!}{r!\,(n - r)!}\). Since each selection can be ordered in \(r!\) ways, \({}^{n}P_{r} = {}^{n}C_{r} \times r!\).
Fig. 1: 'at least 3 women' splits into three separate cases. Within a case multiply (women AND men); across cases add (one case OR another)
Section 01
Order does not matter.
Choosing Ali, Ben and Cy is the same selection as choosing Cy, Ali and Ben. So count the ordered choices \({}^{n}P_{r}\) and divide by the \(r!\) orders of each selection. Check the wording: a committee is a combination, but a chair and a secretary are a permutation.
A team of 5 is chosen from 8 boys and 6 girls. How many teams contain exactly 3 girls?
Choose the girls: \({}^{6}C_{3} = 20\).
Choose the remaining 2 places from the boys: \({}^{8}C_{2} = 28\).
Girls AND boys: \(20 \times 28\).
Answer: 560
Examiner trap
Using \({}^{n}P_{r}\) for a committee. Ordered counting counts each committee \(r!\) times over: \({}^{10}P_{4} = 5040\) is 24 times too big.
Section 02
At least, at most: cases.
'At least' and 'at most' conditions are dealt with by listing every allowed case and adding (Fig. 1). If the list of allowed cases is long but the list of forbidden cases is short, subtract the forbidden cases from the total instead.
Worked example Adding cases
A committee of 5 is chosen from 6 women and 7 men. How many committees contain at least 3 women?
3 women and 2 men: \({}^{6}C_{3} \times {}^{7}C_{2} = 20 \times 21 = 420\).
4 women and 1 man: \({}^{6}C_{4} \times {}^{7}C_{1} = 15 \times 7 = 105\).
5 women: \({}^{6}C_{5} = 6\).
Add the cases: \(420 + 105 + 6\).
Answer: 531
Worked example Using the complement
A group of 4 is chosen from 5 men and 4 women. How many groups contain at least one woman?
Total groups: \({}^{9}C_{4} = 126\).
The only forbidden case is no women, i.e. 4 men: \({}^{5}C_{4} = 5\).
\(126 - 5\).
Answer: 121
Examiner trap
Counting 'at least one woman' as \({}^{4}C_{1} \times {}^{8}C_{3}\): choose one woman, then any 3 of the rest. A group with two women is then counted twice, once for each woman picked 'first'.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Evaluate \({}^{10}C_{4}\).
Quick check+10 XP
Which of these is counted with a combination rather than a permutation?
Quick check+10 XP
Which is equal to \({}^{12}C_{9}\)?
Quick check+10 XP
A group of 4 is chosen from 5 men and 4 women. How many groups contain at least one woman?
Section 03
Particular people.
If a particular person must be included, put them in and choose the rest from those left. If they must be excluded, remove them from the pool. 'Cannot both be chosen' is easiest as total minus 'both chosen'.
Worked example Two who will not serve together
A committee of 5 is chosen from 13 people. Two of them, Mr and Mrs Khan, refuse to serve together. How many committees are possible?
Total: \({}^{13}C_{5} = 1287\).
Both Khans chosen: they fill 2 places, the other 3 come from 11 people: \({}^{11}C_{3} = 165\).
\(1287 - 165\).
Answer: 1122
Worked example One in, one out
From the same 13 people, how many committees of 5 include Mr Khan but not Mrs Khan?
Mr Khan takes one place; Mrs Khan is removed.
Choose the other 4 from the remaining 11: \({}^{11}C_{4}\).
Answer: 330
Your turn
Check the first example another way: count the committees with neither Khan and those with exactly one Khan.
Subtracting only the committees with neither Khan (462). The forbidden case is BOTH Khans, not neither.
Section 04
Groups, then arrangements.
Splitting people into named groups is a chain of selections: choose the first group, then the second from those left. If groups of the same size are unnamed, divide by the number of ways to swap them. When a question selects and then arranges, multiply the combination by the arrangements.
Worked example Groups of different sizes
In how many ways can 9 people be split into groups of 4, 3 and 2?
Choose the 4: \({}^{9}C_{4} = 126\).
Choose the 3 from the 5 left: \({}^{5}C_{3} = 10\). The last 2 form the third group.
\(126 \times 10\).
Answer: 1260
Worked example Equal unnamed groups
In how many ways can 6 people be split into two groups of 3?
Choosing 3 for 'the first group' gives \({}^{6}C_{3} = 20\).
But choosing \(\{A, B, C\}\) or choosing \(\{D, E, F\}\) gives the same split, so each split is counted twice.
Answer: \(20 \div 2 = 10\)
Worked example Select, then arrange
A 4-letter code uses different letters from the word COMPUTER. How many codes include the letter C?
Select: C is in, choose the other 3 letters from the remaining 7: \({}^{7}C_{3} = 35\).
Arrange the 4 chosen letters: \(4! = 24\).
\(35 \times 24 = 840\). Check: all codes \({}^{8}P_{4} = 1680\), minus codes without C, \({}^{7}P_{4} = 840\).
Answer: 840
Examiner trap
Fixing C in the first position (\({}^{7}P_{3} = 210\)). 'Includes C' means C may be in any of the four places.
Stage 2 · Exam
Exam-style questions
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Finish the four checks above to unlock the exam questions
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A committee of 5 is chosen from 6 women and 7 men. How many committees contain at least 3 women?
Exam style+20 XP
A committee of 5 is chosen from 13 people. Two of them will not serve together. How many committees are possible?
Exam style+20 XP
In how many ways can 9 people be divided into groups of 4, 3 and 2?
Exam style+20 XP
A 4-letter code uses different letters chosen from the word COMPUTER. How many codes contain the letter C?
Skill unlocked
Selections, sorted.
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