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Counting
Counting: permutations and combinations
Counting: permutations and combinations Practice
Counting: permutations and combinations Test
CatalogueDPMathematicsMathematics: Analysis and Approaches HLNumber and algebra

Counting: permutations and combinations

Decide whether order changes the outcome before choosing a counting formula.

  • AA HL
  • Papers 1 and 2
  • No GDC needed
  • About 20–25 min

By the end of this lesson

  • use the product rule with or without repetition
  • count ordered selections
  • count unordered selections with restrictions

Understand the idea

Choosing a captain and a deputy is different from choosing a two-person committee. Ada as captain and Ben as deputy is not the same outcome as the reverse; the committee is unchanged when their names swap.

Count choices one position at a time. If every first choice leaves the same number of second choices, multiply those counts. If the remaining choices depend on the case, count the cases separately and add.

Ordered roles: six outcomes

AB, BA, AC, CA, BC, CB

First letter = lead; second = deputy.

Unordered pairs: three outcomes

{A,B}, {A,C}, {B,C}

Each pair occurs in two orders. Divide 6 by 2!.

Six ordered pairs become three unordered pairs.

Checkpoint

Which situation treats swapping two chosen people as a new outcome?
Show explanation

Distinct roles make order meaningful.

Put the idea into notation

From $n$ distinct objects, an ordered selection of $r$ without replacement has $$ {}_nP_r=\frac{n!}{(n-r)!}. $$ With repetition allowed in each of $r$ positions, there are $n^r$ strings.

An unordered selection is counted $r!$ times by the ordered method. Divide out those arrangements: $$ {n\choose r}=\frac{n!}{r!(n-r)!},\quad 0\le r\le n. $$ Here $0!=1$.

For restrictions, fix the required objects first, split into disjoint cases, or subtract forbidden outcomes from the unrestricted total.

Example 1

Assign distinct roles

Choose a lead and an assistant from seven people.

  1. There are 7 choices for lead, then 6 for assistant.
  2. Total $7\times6=42$.
Why does reversing the roles count again?

The same pair with roles reversed is a different assignment.

Example 2

Remove unwanted order

Choose a three-person group from eight people.

  1. Ordered selection gives $8\times7\times6=336$.
  2. Each group appears $3!=6$ times, so $336/6=56$.
Why divide by the number of orders?

All orders of the same three names belong to the same group.

Example 3

Count by the complement

Choose four people from nine, including at least one of two first-aid volunteers.

  1. All groups: ${9\choose4}=126$.
  2. Groups with neither volunteer: ${7\choose4}=35$.
  3. Valid groups: $126-35=91$.
Why count the complement?

Subtracting the complement includes groups with one volunteer and groups with both, without overlap.

In the exam

Quick check

Your score is not saved.

  1. When does choosing two people require counting their two orders separately?
    Show explanation

    Swapping the people changes a role assignment, but not an ordinary two-person group.

  2. How many two-person groups can be selected from five?
    Show explanation

    5 × 4 counts each pair twice, giving 10.

  3. A three-character code uses A, B, C with repetition allowed. A student counts 3 × 2 × 1. What must change?
    Show explanation

    Repetition leaves all three letters available for each position.

  4. A fixed person must be in a group of three chosen from six. How many groups?
    Show explanation

    Choose the other two from five: C(5,2) = 10.

  5. Four distinct talks can run in any order. Talk A must occur before talk B. How many schedules are possible?
    Show explanation

    Pair each of the 4! schedules with the schedule formed by swapping A and B. Exactly one of each pair is allowed.

Lesson Rewards
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Experience points earned for the lesson 5

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