Permutations & combinations drills

Count arrangements and selections using basic permutations and combinations.

Free, timed SHSAT practice sets on permutations & combinations— every question includes a full worked explanation, and there's no account required.

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How to solve permutations & combinations problems

First ask one question: does the order matter?

Common traps: multiplying all the way down to 1 (that arranges everyone, not just the ones you pick), and forgetting to divide when the order does not matter. A quick check: a combination is never larger than the matching permutation.

Worked examples

Example 1.
In how many different ways can 3 of 7 photos be arranged in a line, where the order matters?
Show answer and explanation
Answer:
A

Order matters, so multiply 3 factors counting down from 7: 7 \(\times\) 6 \(\times\) 5 = 210. (Arranging all 7 photos (\(7!\)) gives 5040. Dividing by \(3!\) as if order did not matter gives 35. Letting the same photo be picked more than once (\(7^{3}\)) gives 343.)

Example 2.
How many different 5-person committees can be chosen from a group of 7 candidates, where the order of selection does not matter?
Show answer and explanation
Answer:
A

Order does not matter, so divide the ordered count by \(5!\): \((7 \times 6 \times 5 \times 4 \times 3) \div 5! = 2520 \div 120 = 21\). (Not dividing by \(5!\) (\(7 \times 6 \times 5 \times 4 \times 3\)) gives 2520. Dividing by 5 instead of \(5! = 120\) gives 504. Dividing \(7!\) by \(5!\) only (\(7! \div 5!\)) gives 42.)

Example 3.
How many different 4-person committees can be chosen from a group of 11 volunteers, where the order of selection does not matter?
Show answer and explanation
Answer:
B

Order does not matter, so divide the ordered count by \(4!\): \((11 \times 10 \times 9 \times 8) \div 4! = 7920 \div 24 = 330\). (Not dividing by \(4!\) (\(11 \times 10 \times 9 \times 8\)) gives 7920. Dividing by 4 instead of \(4! = 24\) gives 1980. Dividing \(11!\) by \(4!\) only (\(11! \div 4!\)) gives 1663200.)