IGCSE · Practice questions · Arrangement and selection problems

Read the verbs, then count.

Six original Cambridge-style questions on arrangement and selection: deciding between permutations and combinations, digit problems without repetition, at least and at most, and algebraic equations in \(n\).

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

Decide, split, count, check.

Decide: order matters (arranged, positions, numbers) gives \({}^{n}P_{r}\); order does not (chosen, committee, team) gives \({}^{n}C_{r}\).
Restricted places first: in digit problems fill the last digit (even, divisible by 5) and the first digit (size, no leading 0) before the rest.
Split into cases: 'and' multiplies, 'or' adds, and cases must not overlap.
At least one: total minus none is usually quickest.
Find n: write out the formula, cancel, solve, reject negative or too-small \(n\), and check by substituting.

Recall

[2 marks]

A club has 12 members. Find the number of ways of

(a) choosing a president, a secretary and a treasurer,

(b) choosing a committee of 3 members.

(a) Three different roles, so order matters: \({}^{12}P_{3} = 12 \times 11 \times 10 = 1320\). ✓
(b) A committee has no order: \({}^{12}C_{3} = \dfrac{1320}{3!} = 220\). ✓

Calculation

[3 marks]

Four-digit numbers are formed from the digits 0, 1, 2, 3, 4, 5, with no digit used more than once. Find how many of these numbers are divisible by 5.

Divisible by 5 means the last digit is 0 or 5. ✓
Last digit 0: the first three places from the other 5 digits, \({}^{5}P_{3} = 60\). ✓
Last digit 5: the first digit cannot be 0, so it is 1, 2, 3 or 4 (4 ways); the middle two from the 4 digits left, \({}^{4}P_{2} = 12\); \(4 \times 12 = 48\). ✓
Add the two cases: \(60 + 48 = 108\). ✓

Calculation

[4 marks]

Four-digit numbers are formed from the digits 1, 2, 3, 4, 5, 6, 7, with no digit used more than once. Find how many of these numbers are odd and greater than 4000.

The first digit is 4, 5, 6 or 7 and the last digit is 1, 3, 5 or 7. Split by the first digit. ✓
First digit 4 or 6 (even): last digit any of the 4 odd digits; middle two from the 5 digits left, \({}^{5}P_{2} = 20\). Count \(2 \times 4 \times 20 = 160\). ✓
First digit 5 or 7 (odd): only 3 odd digits are left for the end; middle \({}^{5}P_{2} = 20\). Count \(2 \times 3 \times 20 = 120\). ✓
Total \(160 + 120 = 280\). ✓

Analysis

[5 marks]

A choir has 7 sopranos and 5 altos. A group of 6 singers is chosen.

(a) Find the number of groups that contain at least 4 sopranos.

(b) Two of the sopranos, Hana and Isla, refuse to sing in the same group. Find the number of groups of 6 that can be chosen from the whole choir with no other restriction.

(a) 4S 2A: \({}^{7}C_{4} \times {}^{5}C_{2} = 35 \times 10 = 350\). ✓
5S 1A: \({}^{7}C_{5} \times {}^{5}C_{1} = 21 \times 5 = 105\). 6S: \({}^{7}C_{6} = 7\). ✓
Total \(350 + 105 + 7 = 462\). ✓
(b) All groups: \({}^{12}C_{6} = 924\). Groups with both Hana and Isla: choose 4 more from 10, \({}^{10}C_{4} = 210\). ✓
Allowed groups: \(924 - 210 = 714\). ✓

Analysis

[4 marks]

(a) Given that \({}^{n}C_{3} = 2 \times {}^{n}C_{2}\), find the value of \(n\).

(b) Given that \({}^{n+2}P_{2} = 110\), find the value of \(n\).

(a) \(\dfrac{n(n - 1)(n - 2)}{6} = 2 \times \dfrac{n(n - 1)}{2} = n(n - 1)\). ✓
Divide by \(n(n - 1)\), which is not zero since \(n \ge 3\): \(\dfrac{n - 2}{6} = 1\), so \(n = 8\). Check: \({}^{8}C_{3} = 56 = 2 \times 28\). ✓
(b) \((n + 2)(n + 1) = 110 = 11 \times 10\), so \(n^{2} + 3n - 108 = 0\). ✓
\((n + 12)(n - 9) = 0\); reject \(n = -12\), so \(n = 9\). Check: \({}^{11}P_{2} = 11 \times 10 = 110\). ✓

Analysis

[6 marks]

The six letters of the word FACTOR are written on six cards.

(a) Find the number of different arrangements of all six cards in a row.

(b) Find the number of these arrangements that begin with a vowel.

(d) Three of the six cards are chosen. Find the number of selections that include at least one vowel.

(a) Six different letters in a row: \(6! = 720\). ✓
(b) First letter A or O (2 ways), then the other 5 in any order: \(2 \times 5! = 240\). ✓
(c) The consonants are F, C, T, R. First letter 4 ways, last letter 3 ways, middle four letters \(4! = 24\): \(4 \times 3 \times 24 = 288\). ✓
(d) A selection has no order. All selections \({}^{6}C_{3} = 20\); with no vowel (3 of the 4 consonants) \({}^{4}C_{3} = 4\). ✓
At least one vowel: \(20 - 4 = 16\). ✓

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.