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Year 8 · Statistics & Probability

Counting arrangements

Counting arrangements: n! orders of n things; permutations when order matters; combinations when it does not.

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How to do it

  1. n! = n × (n − 1) × … × 1.
  2. nPr = n! ÷ (n − r)!
  3. nCr = n! ÷ (r! (n − r)!)

Worked example

Evaluate 6C3 (combinations of 3 from 6).
nCr = n! ÷ (r!(n − r)!) = 120 ÷ 3! = 20.
Answer: 20

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Teams and groups: combinations. Rankings and codes: permutations.

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