IGCSE Additional Mathematics 0606 · Topic 12

IGCSE Additional Mathematics: Permutations and Combinations Practice Questions

A permutation counts arrangements where order matters, and a combination counts selections where it does not. Deciding which applies is the whole question, and the word chosen or selected almost always signals a combination.

Cambridge IGCSE Additional Mathematics (0606) · Topic 12: Permutations and Combinations

Topic 12 of Cambridge IGCSE Additional Mathematics 0606 is self contained and quick to master. The at least one condition is best handled by subtracting from the total. The questions below cover arrangements, selections, repeated letters and that subtraction.

What you need to know for Permutations and Combinations

IGCSE Additional Mathematics Permutations and Combinations questions and answers

4 exam-style questions written to the 0606 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.

Question 1[3 marks]

In how many different orders can 5 different books be arranged on a shelf?

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Answer: 120
  1. Order matters, so this is a permutation of all 5 items.
  2. The number of arrangements is 5 factorial.
  3. 5! = 5 x 4 x 3 x 2 x 1 = 120.
How the marks are awarded. 1 mark for recognising this as an arrangement. 1 mark for using 5 factorial. 1 mark for 120.
Where students lose the mark. Using a combination. The books are being placed in order on a shelf, so different orders count as different outcomes.
Question 2[4 marks]

From a group of 8 people, in how many ways can a team of 3 be chosen, and in how many ways can 3 be arranged in first, second and third place?

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Answer: 56 teams and 336 arrangements
  1. Choosing a team means order does not matter, so use a combination: 8 choose 3.
  2. 8 choose 3 = 8 x 7 x 6 divided by (3 x 2 x 1) = 336 divided by 6 = 56.
  3. Awarding first, second and third place means order matters, so use a permutation: 8 permute 3.
  4. 8 permute 3 = 8 x 7 x 6 = 336. Note the permutation is 6 times the combination, because each team of 3 can be ordered in 3! = 6 ways.
How the marks are awarded. 1 mark for identifying a combination for the team. 1 mark for 56. 1 mark for identifying a permutation for the placings. 1 mark for 336.
Where students lose the mark. Using the same calculation for both parts. The difference between the two answers is exactly the 3 factorial orderings of each selected group.
Question 3[4 marks]

How many different arrangements can be made of all the letters of the word ARRANGE?

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Answer: 1260
  1. ARRANGE has 7 letters, so without repeats there would be 7! = 5040 arrangements.
  2. The letter A appears twice and the letter R appears twice.
  3. Swapping the two identical A letters produces no new arrangement, and the same applies to the two R letters.
  4. Divide by 2! for the A letters and 2! for the R letters: 5040 divided by (2 x 2) = 1260.
How the marks are awarded. 1 mark for 7 factorial. 1 mark for identifying both repeated letters. 1 mark for dividing by 2! twice. 1 mark for 1260.
Where students lose the mark. Dividing by 2 only once. Each repeated letter needs its own division, so two repeated pairs mean dividing by 2! twice.
Question 4[4 marks]

A committee of 3 is chosen from 5 men and 4 women. In how many ways can the committee be formed if it must contain at least one woman?

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Answer: 74
  1. Total number of committees of 3 from 9 people: 9 choose 3 = 84.
  2. Count the committees with no women, meaning all 3 chosen from the 5 men: 5 choose 3 = 10.
  3. At least one woman is the complement of no women.
  4. 84 minus 10 = 74 committees.
How the marks are awarded. 1 mark for a total of 84. 1 mark for identifying the all men case. 1 mark for 10. 1 mark for 74.
Where students lose the mark. Adding the cases for exactly one, two and three women and missing one. Subtracting from the total is faster and safer.

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Permutations and Combinations FAQs

What is the difference between a permutation and a combination?

A permutation counts arrangements where the order matters, such as placings in a race or letters in a code. A combination counts selections where order does not matter, such as a team or a committee. The permutation count is always the larger of the two.

How do I count arrangements of a word with repeated letters?

Start with the factorial of the total number of letters, then divide by the factorial of the count of each repeated letter. The word ARRANGE has 7 letters with two A letters and two R letters, giving 7 factorial divided by 2 factorial twice, which is 1260.

How do I handle at least one conditions?

Count the total number of possibilities, then count the cases containing none of the required item, and subtract. Counting each qualifying case directly takes longer and risks omitting one, which is why the subtraction method is preferred.

How do I decide which method a question needs?

Ask whether swapping two of the chosen items produces a different outcome. If it does, order matters and you need a permutation. If it does not, order is irrelevant and you need a combination. Words such as team, committee and selection signal combinations.

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Written to the published Cambridge IGCSE Additional Mathematics (0606) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.