About Permutation and Combination
The whole topic turns on one question: does the order matter? If it does the answer is a permutation, and if it does not the answer is a combination. Arrangements of letters, selections of committees, prize distributions and seating plans are all one or other of those two.
What you need to understand
- A permutation counts arrangements, where the order matters. A combination counts selections, where it does not.
- The number of ways to arrange r things out of n is nPr = n! / (n - r)!.
- The number of ways to choose r things out of n is nCr = n! / (r! (n - r)!).
- A combination is always smaller than the matching permutation by a factor of r!, because each selection can be arranged in r! ways.
- The number of ways to arrange n distinct letters is n!.
- If some letters repeat, divide n! by the factorial of each repeat count, because swapping two identical letters does not create a new arrangement.
Formulas to remember
How to work through these questions
- Ask whether swapping two of the chosen items gives a different outcome. If it does, use a permutation; if it does not, use a combination.
- For a committee, a team or a hand of cards use a combination, because the order of selection is never recorded.
- For prizes, posts, seating or passwords use a permutation, because the positions are distinct.
- For a word, count the letters first and then list the repeated letters with their counts.
- Check the answer against n!, because no arrangement can exceed the total number of orderings of all the letters.
Mistakes that cost marks
- Using a permutation where only a selection is required, which inflates the answer by a factor of r!.
- Using a combination where the order matters, which shrinks the answer instead.
- Forgetting to divide by the factorials of the repeated letters in a word.
- Treating nCr as n multiplied by r.
- Allowing the same item to be chosen twice when the question says each may be used at most once.
Worked example
In how many ways can a committee of 3 be selected from 8 people?
- The order of selection does not matter, so this is a combination rather than a permutation.
- 8C3 = 8! / (3! x 5!) = (8 x 7 x 6) / (3 x 2 x 1).
- The numerator is 336 and the denominator is 6, so the answer is 56.
- Check: arranging 3 of the 8 people would give 8 x 7 x 6 = 336 ways, which is 56 x 6, exactly six times as many as the selections.
Answer: 56
Practice questions with answers
A few Permutation and Combination questions with the full solution shown, so you can see
how the method is applied before you attempt the timed set.
Question 1
Out of 12 students, 2 are to be chosen for a quiz team. In how many ways can the team be chosen?
Answer: Option C — with explanation
The order within the group does not matter, so this is a combination.
12C2 = 12! / (2! x 10!) = 66.
Common mistakes
- allowed the same person to be chosen repeatedly: 144
- left one person out of the group altogether: 55
- arranged the chosen members instead of only selecting them: 132
Question 2
In how many ways can 2 distinct posts be filled by 9 candidates, no candidate holding more than one post?
Answer: Option B — with explanation
The prizes are different, so the order matters and this is a permutation.
9P2 = 9! / (9 - 2)! = 72.
Common mistakes
- allowed one student to receive more than one prize: 81
- selected the students without allowing for which prize each one receives: 36
- handed out one prize fewer: 9
Question 3
In how many ways can 3 different prizes be given to 8 students if no student may receive more than one prize?
Answer: Option D — with explanation
The prizes are different, so the order matters and this is a permutation.
8P3 = 8! / (8 - 3)! = 336.
Common mistakes
- allowed one student to receive more than one prize: 512
- multiplied the two numbers: 24
- selected the students without allowing for which prize each one receives: 56
Question 4
How many different groups of 5 can be formed from 9 people?
-
A
15120
-
B
59049
-
C
45
-
D
126
Answer: Option D — with explanation
The order within the group does not matter, so this is a combination.
9C5 = 9! / (5! x 4!) = 126.
Common mistakes
- arranged the chosen members instead of only selecting them: 15120
- multiplied the two numbers instead of choosing: 45
- allowed the same person to be chosen repeatedly: 59049
Question 5
In how many ways can a committee of 4 members be selected from 9 people?
Answer: Option D — with explanation
The order within the group does not matter, so this is a combination.
9C4 = 9! / (4! x 5!) = 126.
Common mistakes
- arranged the chosen members instead of only selecting them: 3024
- allowed the same person to be chosen repeatedly: 6561
- multiplied the two numbers instead of choosing: 36
Frequently asked questions
How do I decide between a permutation and a combination?
Ask whether a different order is a different outcome. Selecting a committee of three is a combination, while awarding first, second and third prizes to three people is a permutation, because the prizes are different.
Why is nCr equal to nPr divided by r!?
Because any group of r items can be ordered in r! different ways, and a combination counts all of those orderings as a single choice. Dividing removes the over-counting.
How do I handle a word with repeated letters?
Divide n! by the factorial of each repeat count. The word BANANA has six letters with A appearing three times and N twice, so the count is 720 / (6 x 2) = 60.
How many ways are there to seat people around a circular table?
Use (n - 1)!, because rotating everybody round the table does not produce a new arrangement. For five people that is 4! = 24.
