AS & A Level · Practice questions · Combinations: selections

Cases, then add.

Six original Cambridge-style questions on combinations and selections with conditions, at Paper 5 standard.

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

The method

Read the condition, then choose a strategy.

Order irrelevant: \({}^{n}C_{r}\); order relevant: \({}^{n}P_{r} = {}^{n}C_{r} \times r!\).
From two groups: multiply the choices from each group.
At least / at most: list the cases and add, or total minus forbidden cases.
Particular people: include them and choose the rest, or remove them from the pool.
Equal unnamed groups: divide by the number of ways to swap them.

Recall

[2 marks]

Find the values of \({}^{8}C_{3}\) and \({}^{8}P_{3}\), and explain why one is 6 times the other.

\({}^{8}C_{3} = 56\), \({}^{8}P_{3} = 336\). ✓
Each selection of 3 objects can be arranged in \(3! = 6\) orders, so there are 6 times as many ordered choices. ✓

Calculation

[4 marks]

A team of 6 is chosen from 7 boys and 5 girls. Find the number of teams containing (a) exactly 2 girls, (b) at least 2 girls.

(a) \({}^{5}C_{2} \times {}^{7}C_{4} = 10 \times 35 = 350\). ✓
(b) Total \({}^{12}C_{6} = 924\). No girls: \({}^{7}C_{6} = 7\). One girl: \({}^{5}C_{1} \times {}^{7}C_{5} = 105\). ✓
\(924 - 7 - 105 = 812\). ✓

Calculation

[2 marks]

A committee of 5 is chosen from 12 people, including Mr and Mrs Lee, who refuse to serve together. Find the number of possible committees.

Total \({}^{12}C_{5} = 792\); both Lees: \({}^{10}C_{3} = 120\). ✓
\(792 - 120 = 672\). ✓

Calculation

[3 marks]

Ten students are divided into groups. Find the number of ways of dividing them into (a) groups of 5, 3 and 2, (b) two groups of 5.

(a) \({}^{10}C_{5} \times {}^{5}C_{3} = 252 \times 10 = 2520\). ✓
(b) \({}^{10}C_{5} = 252\) counts each split twice (the group chosen and the group left behind can be swapped), so \(252 \div 2 = 126\). ✓

Analysis

[4 marks]

Find the number of different selections of 4 letters that can be made from the 8 letters of the word PARALLEL.

The letters are L (3 times), A (twice), and P, R, E once each. Split by the number of Ls. ✓
Three Ls: the fourth letter is A, P, R or E: 4 selections. ✓
Two Ls: AA (1), or two different letters from A, P, R, E: \({}^{4}C_{2} = 6\). Total 7. ✓
One L: AA with one of P, R, E (3), or three different from A, P, R, E: \({}^{4}C_{3} = 4\). Total 7. ✓
No L: AA with two of P, R, E (3), or A, P, R, E (1). Total 4. ✓
Altogether \(4 + 7 + 7 + 4 = 22\). ✓

Analysis

[5 marks]

A choir of 7 singers is chosen from 6 sopranos, 5 altos and 4 tenors. The choir must contain at least 2 singers of each type.

(a) Find the number of different choirs. (b) If instead the 7 singers were chosen at random from the 15, find the probability that the condition in (a) would be met.

(a) Allowed cases (S, A, T): (3, 2, 2), (2, 3, 2), (2, 2, 3). ✓
\({}^{6}C_{3}\,{}^{5}C_{2}\,{}^{4}C_{2} = 20 \times 10 \times 6 = 1200\). ✓
\({}^{6}C_{2}\,{}^{5}C_{3}\,{}^{4}C_{2} = 15 \times 10 \times 6 = 900\). ✓
\({}^{6}C_{2}\,{}^{5}C_{2}\,{}^{4}C_{3} = 15 \times 10 \times 4 = 600\). Total 2700. ✓
(b) All choirs: \({}^{15}C_{7} = 6435\). Probability \(= \dfrac{2700}{6435} = 0.420\) (3 s.f.). ✓

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.