IGCSE · Practice questions · Factorials and permutations

Restrictions first, the rest freely.

Six original Cambridge-style questions on factorials and permutations: evaluating \(n!\) and \({}^{n}P_{r}\), solving a permutation equation, forming numbers from digits without repetition and arranging people and letters with items together, apart or in fixed places.

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

One box per place, restricted boxes first.

Draw a box for each position and write the number of choices left in it; multiply.
All \(n\) different items: \(n!\). Only \(r\) of them, in order: \({}^{n}P_{r} = \dfrac{n!}{(n - r)!}\).
Restricted boxes first: the last digit for even/odd, the first digit for size (never 0), fixed ends.
Together: one block, then multiply by the arrangements inside. Apart (two items): total minus together.
Both ends restricted by the same digits or letters? Split into cases and add.

Recall

[2 marks]

Without using a calculator, find the value of (a) \({}^{8}P_{3}\) (b) \(\dfrac{12!}{10! \times 2!}\).

(a) \({}^{8}P_{3} = 8 \times 7 \times 6 = 336\). ✓
(b) \(\dfrac{12 \times 11 \times 10!}{10! \times 2} = \dfrac{132}{2} = 66\). ✓

Calculation

[4 marks]

Three-digit numbers are formed from the digits 0, 1, 2, 3, 4, 5. No digit may be used more than once, and a number cannot begin with 0.

(a) How many such numbers are there?

(b) How many of them are even?

(a) First digit: any of 1 to 5, so 5 ways. Then 5 digits are left (including 0) for the second place and 4 for the third: \(5 \times 5 \times 4 = 100\). ✓
(b) Split on the last digit, because 0 changes what the first digit can be. ✓
Last digit 0: first digit 5 ways, middle 4 ways: \(5 \times 4 = 20\). ✓
Last digit 2 or 4 (2 ways): the first digit cannot be 0 or the last digit, so 4 ways; middle 4 ways: \(2 \times 4 \times 4 = 32\). ✓
Even numbers: \(20 + 32 = 52\). ✓

Calculation

[3 marks]

(a) Show that \(\dfrac{n!}{(n - 2)!} = n^{2} - n\).

(b) Hence solve \({}^{n}P_{2} = 132\).

(a) \(n! = n(n - 1)(n - 2)!\), so \(\dfrac{n!}{(n - 2)!} = n(n - 1) = n^{2} - n\). ✓
(b) \(n^{2} - n = 132 \Rightarrow n^{2} - n - 132 = 0 \Rightarrow (n - 12)(n + 11) = 0\). ✓
\(n\) is a positive integer, so \(n = 12\) (reject \(-11\)). ✓

Calculation

[4 marks]

Eight friends, including Kofi and Lena, stand in a row for a photograph.

(a) In how many ways can they stand?

(b) In how many of these arrangements are Kofi and Lena next to each other?

(a) \(8! = 40\,320\). ✓
(b) Block [KL] plus 6 others: \(7! = 5040\) arrangements, times \(2!\) for KL or LK: \(10\,080\). ✓
(c) Total minus together: \(40\,320 - 10\,080 = 30\,240\). ✓

Analysis

[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 5000.

The first digit must be 5, 6 or 7, and the last digit must be odd. Both restrictions involve 5 and 7, so split on the first digit. ✓
First digit 5 or 7 (2 ways): the odd digits left for the end are 3 of 1, 3, 5, 7 (3 ways); middle \(5 \times 4 = 20\): \(2 \times 3 \times 20 = 120\). ✓
First digit 6 (1 way): the last digit is any of 1, 3, 5, 7 (4 ways); middle \(5 \times 4 = 20\): \(1 \times 4 \times 20 = 80\). ✓
Total \(120 + 80 = 200\). ✓

Analysis

[6 marks]

The 8 letters of the word EQUATION are all different. They are arranged in a row.

(a) Find the total number of arrangements.

(b) Find the number of arrangements that begin with a consonant and end with a vowel.

(d) Find the number of arrangements in which Q and N are not next to each other.

(a) \(8! = 40\,320\). ✓
(b) Consonants Q, T, N (3); vowels E, U, A, I, O (5). First box 3 ways, last box 5 ways, the middle 6 letters \(6! = 720\) ways: \(3 \times 5 \times 720 = 10\,800\). ✓
(c) Block [QTN] plus the 5 vowels: \(6! = 720\) arrangements, times \(3! = 6\) inside the block: \(4320\). ✓
(d) Q and N together: block [QN] plus 6 other letters, \(7! \times 2! = 10\,080\). ✓
Not together: \(40\,320 - 10\,080 = 30\,240\). ✓

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.