IGCSE · Practice questions · Combinations

Choose, multiply, subtract.

Six original Cambridge-style questions on combinations: deciding whether order matters, using \({}^{n}C_{r}\) and its symmetry, solving equations in \(n\), and selecting committees from groups with 'at least' conditions and named people, all without a calculator.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Does order matter? Then build the count.

Decide: named roles, prizes or positions mean order matters (\(P\)); a plain group means it does not (\(C\)).
Evaluate \({}^{n}C_{r}\) by cancelling: \(\dfrac{n(n - 1)\cdots(n - r + 1)}{r!}\), using \({}^{n}C_{r} = {}^{n}C_{n-r}\) to keep \(r\) small.
Groups: choose from each group and multiply; add the separate cases.
At least one: total minus none. Must be in: place them, choose the rest from the others. Not both: total minus both.
Equations: write \({}^{n}C_{2} = \dfrac{n(n - 1)}{2}\), solve the quadratic and reject negative \(n\).

Recall

[2 marks]

Evaluate \({}^{9}C_{4}\) and \({}^{9}C_{5}\), and explain why they are equal.

\({}^{9}C_{4} = \dfrac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \dfrac{3024}{24} = 126\). ✓
Choosing 4 of 9 to take is the same as choosing the 5 to leave, so \({}^{9}C_{5} = {}^{9}C_{4} = 126\). ✓

Calculation

[3 marks]

There are 11 athletes in a club.

(a) In how many ways can a squad of 4 athletes be selected for a competition?

(b) In how many ways can first, second, third and fourth places be awarded to 4 of the 11 athletes in a race?

(a) Order does not matter: \({}^{11}C_{4} = \dfrac{11 \times 10 \times 9 \times 8}{24} = 330\). ✓
(b) The places are different, so order matters: \({}^{11}P_{4} = 11 \times 10 \times 9 \times 8 = 7920\). ✓
(c) Each squad of 4 from (a) can be given the 4 places in \(4! = 24\) different orders, and \(330 \times 24 = 7920\). ✓

Calculation

[3 marks]

Given that \({}^{n}C_{2} = 105\), find the value of \(n\).

\(\dfrac{n(n - 1)}{2} = 105\), so \(n^{2} - n - 210 = 0\). ✓
\((n - 15)(n + 14) = 0\), so \(n = 15\) or \(n = -14\). ✓
\(n\) must be a positive integer, so \(n = 15\). Check: \(\dfrac{15 \times 14}{2} = 105\). ✓

Calculation

[4 marks]

A committee of 6 people is to be chosen from 7 men and 5 women. Find the number of different committees that contain

(a) exactly 4 men,

(b) at least one woman.

(a) 4 men and 2 women: \({}^{7}C_{4} \times {}^{5}C_{2} = 35 \times 10 = 350\). ✓
(b) Total committees: \({}^{12}C_{6} = 924\). ✓
Committees with no woman: \({}^{7}C_{6} = 7\). ✓
At least one woman: \(924 - 7 = 917\). ✓

Analysis

[5 marks]

Five students are to be chosen from a class of 12 to form a debating team. Ali and Ben are two of the 12 students. Find the number of different teams that

(a) include Ali,

(b) do not include both Ali and Ben,

(a) Put Ali in the team and choose the other 4 from the other 11: \({}^{11}C_{4} = 330\). ✓
(b) Total teams \({}^{12}C_{5} = 792\). Teams with both: choose 3 more from the other 10, \({}^{10}C_{3} = 120\). ✓
Not both: \(792 - 120 = 672\). ✓
(c) Choose which one of Ali and Ben (2 ways), leave the other out, then choose 4 from the other 10: \(2 \times {}^{10}C_{4} = 2 \times 210 = 420\). ✓
Check: (b) is 'exactly one' plus 'neither': \(420 + {}^{10}C_{5} = 420 + 252 = 672\). ✓

Analysis

[6 marks]

A school has \(n\) staff. The number of ways of choosing 3 of them is 4 times the number of ways of choosing 2 of them.

(a) Show that \(n = 14\).

(b) The 14 staff are 8 teachers and 6 support staff. A working group of 4 is chosen from them. Find the number of groups that contain at least 2 support staff.

(a) \({}^{n}C_{3} = 4 \times {}^{n}C_{2}\) gives \(\dfrac{n(n - 1)(n - 2)}{6} = 4 \times \dfrac{n(n - 1)}{2}\). ✓
\(n \ge 3\), so \(n(n - 1) \ne 0\); divide both sides by \(n(n - 1)\): \(\dfrac{n - 2}{6} = 2\), so \(n - 2 = 12\) and \(n = 14\). ✓
(b) 2 support staff: \({}^{6}C_{2} \times {}^{8}C_{2} = 15 \times 28 = 420\). ✓
3 support staff: \({}^{6}C_{3} \times {}^{8}C_{1} = 20 \times 8 = 160\); 4 support staff: \({}^{6}C_{4} = 15\). ✓
Total: \(420 + 160 + 15 = 595\). ✓
Check: \({}^{14}C_{4} = 1001\); fewer than 2 support staff is \({}^{8}C_{4} + 6 \times {}^{8}C_{3} = 70 + 336 = 406\), and \(1001 - 406 = 595\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.