Permutations and combinations: notes and questions
Revision for arrangements and selections of different objects, with and without restrictions. Cambridge 0606 only. Repeated objects and arrangements in a circle are not on the syllabus.
Objects that must stay together: treat them as one block, then multiply by the arrangements inside the block.
'At least one' is usually easiest as total minus none.
When a condition splits the problem (more girls than boys, first digit 5 or 6), list the cases and add.
Watch out: Deal with the restricted positions first (ends, first digit, last digit), then fill the rest.
Worked example· 1 mark
Find the number of different arrangements of the six letters of the word PLANET.
B1 6! = 720
All six letters are different, so there are 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 arrangements.
Answer: \(720\)
Permutations and combinations questions
Original exam-style questions, written for this site and each re-solved independently. Tags show the course whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).
Question 1 · 2 marks0606no calculator
How many arrangements of the letters of the word PLANET begin with P and end with T?
Show the mark scheme and worked solution
M1 P and T fixed, 4 letters arranged
A1 24
P and T are fixed at the ends.
The other 4 letters can be arranged in 4! = 24 ways.
Answer: \(24\)
Question 2 · 2 marks0606no calculator
A committee of 4 people is chosen from 7 men and 5 women. How many committees contain at least one woman?
Show the mark scheme and worked solution
M1 total − all men: ¹²C₄ − ⁷C₄
A1 460
Total committees: ¹²C₄ = 495.
Committees with no women: ⁷C₄ = 35.
At least one woman: 495 − 35 = 460.
Answer: \(460\)
Question 3 · 2 marks0606no calculator
Find the value of \(n\) for which \(^{n}C_{2} = 45\).
Show the mark scheme and worked solution
M1 n(n − 1)/2 = 45
A1 n = 10
ⁿC₂ = n(n − 1)/2 = 45, so n² − n − 90 = 0.
(n − 10)(n + 9) = 0, and n must be positive, so n = 10.
Answer: \(n = 10\)
Question 4 · 3 marks0606no calculator
Three different maths books and four different science books are placed in a row. The three maths books must be next to each other. In how many ways can this be done?
Show the mark scheme and worked solution
M1 5! for the block and 4 science books
M1 × 3! inside the block
A1 720
Treat the maths books as one block: the block and 4 science books make 5 items, in 5! = 120 orders.
The maths books can be arranged inside the block in 3! = 6 ways.
120 × 6 = 720.
Answer: \(720\)
9 more permutations and combinations questions
Each with a mark scheme and a worked solution. Included with IGCSE plans, free trials and school licences.
IGCSE Math Revision