Counting Principles (DP IB Analysis & Approaches (AA)): Revision Note

Counting principles

How do I count possibilities?

• The fundamental multiplication rule is

  • if there are possible choices

  • followed by possible choices

  • then there are possible choices in total

• e.g. if a menu has 5 starters and 6 desserts

  • then there are 30 possible combinations of a starter and a dessert

How do I count with digits?

• Recall that there are 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

• These can be used as choices when counting

  • e.g. how many 3-digit PIN numbers are there?

    • 10 choices for the first digit

    • 10 choices for the second digit

    • 10 choices for the third digit

    • 10 × 10 × 10 = 1000 possible 3-digit PIN numbers

How do I count with no repeats allowed?

• This is best shown through an example

  • e.g. how many 3-digit PIN numbers are there with no repeated digits?

    • 10 choices for the first digit

    • 9 remaining choices for the second digit

    • 8 remaining choices for the third digit

    • 10 × 9 × 8 = 720 possible 3-digit PIN numbers with no repeats

How do I count with restrictions?

• This is best shown through an example

• e.g. how many 3-digit PIN numbers have an odd first digit and a square third digit?

  • There are 5 choices for the first digit

    • 1, 3, 5, 7, 9

  • There are 10 choices for the second digit

    • no restrictions on this one

  • There are 4 choices for the third digit

    • 0 (= 0²), 1 (= 1²), 4 (= 2²), 9 (= 3²)

  • So there are 5 × 10 × 4 = 200 possibilities

How do I count using cases?

•  Sometimes splitting into cases can make the question easier

  • Calculate the separate possibilities

    • then add the results from each case together

• e.g. how many 3-digit PIN codes start with either 3 or a 7 and end in an odd digit?

  • Case 1: it starts with a 3

    • There is 1 possible choice for the first digit (a 3)

    • There are 10 possible choices for the second digit

    • There are 5 possible choices for the third digit

    • So 1 × 10 × 5 = 50 possibilities in case 1

  • Case 2: it starts with a 7

    • There is 1 possible choice for the first digit (a 7)

    • There are 10 possible choices for the second digit

    • There are 5 possible choices for the third digit

    • So 1 × 10 × 5 = 50 possibilities in case 2

  • The total number of possibilities is 50 + 50 = 100