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Distinguish permutations from combinations

The distinction is that permutations count ordered arrangements, whereas combinations count unordered selections; for nn distinct objects chosen rr at a time without repetition, these are represented by nPr=n!(nr)!nP r=\frac{n!}{(n-r)!} and nCr=n!r!(nr)!nC r=\frac{n!}{r!(n-r)!}, respectively. The learner understands that treating equivalent groups as different overcounts combinations, and applies the distinction to finite counting and elementary probability situations, without extending to advanced generalized or repeated-selection cases.

Detailed Explanation: Distinguish permutations from combinations

When counting, first ask:

Does the order of the objects matter?

  • If different orders count as different outcomes, use a permutation:
nPr=n!(nr)! nPr=\frac{n!}{(n-r)!}
  • If different orders represent the same group, use a combination:
nCr=n!r!(nr)! nCr=\frac{n!}{r!(n-r)!}

Here, nn is the total number of distinct objects, and rr is the number chosen.

Example

There are 66 students: Alice, Ben, Carlos, Dana, Emma, and Farah.

Part 1: Choose 3 students for a committee

The committee {Alice, Ben, Carlos}\{\text{Alice, Ben, Carlos}\} is the same group as {Carlos, Alice, Ben}\{\text{Carlos, Alice, Ben}\}. Therefore, order does not matter, so use a combination:

6C3=6!3!(63)!6C3=\frac{6!}{3!(6-3)!} 6C3=6!3!3!=72066=206C3=\frac{6!}{3!3!} =\frac{720}{6\cdot 6} =20

There are 20\boxed{20} possible committees.

The factor (3!)(3!) in the denominator removes the overcounting caused by arranging the same three students in different orders.

Part 2: Choose a president, vice president, and secretary

Now the roles are different. For example, Alice as president and Ben as vice president is different from Ben as president and Alice as vice president. Therefore, order matters, so use a permutation:

6P3=6!(63)!6P3=\frac{6!}{(6-3)!} 6P3=6!3!=654=1206P3=\frac{6!}{3!} =6\cdot5\cdot4 =120

There are 120\boxed{120} possible assignments.

A useful check is that each group of 33 students can be assigned to the three roles in (3!)(3!) ways:

203!=206=12020\cdot 3!=20\cdot6=120

So, use a combination for an unordered group and a permutation when the positions or roles make the order important.

Learn by doing: Distinguish permutations from combinations

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Probability Permutation or Combination - Scenario to Value


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