AQA · Mathematics · 8300

GCSE Maths: Product rule for counting

Probability · AQA GCSE Maths (8300). Cards are generated from the Corbettmaths practice questions for Product rule for counting (Corbettmaths Video 383), and every card links back to the exact page it came from. Practice tests are drawn from this deck’s own cards. Source material © Corbettmaths — used with attribution.

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  1. What does the product rule for counting state?

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    If one choice can be made in m ways and a second independent choice in n ways, the two together can be made in m × n ways.

  2. How many ways can one of 20 Year 4 students and one of 18 Year 5 students be chosen?

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    20 × 18 = 360 ways.

  3. How many football strips can be made from 4 pairs of socks, 3 pairs of shorts and 11 shirts?

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    4 × 3 × 11 = 132 strips.

  4. 16 rugs and some mirrors give 144 rug-and-mirror combinations. How many mirrors are there?

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    144 ÷ 16 = 9 mirrors (reverse the product rule by dividing).

  5. A 3-dial lock uses digits 1–8 on each dial. How many settings have three different digits?

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    8 × 7 × 6 = 336.

  6. Why is 8 × 7 × 6 used, not 8³, when three chosen digits must all be different?

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    Each digit used removes one option for the next choice: 8 options, then 7, then 6 (no repetition).

  7. How many 4-digit codes start with 2 and are odd (digits 0–9, repeats allowed)?

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    1 × 10 × 10 × 5 = 500 (last digit must be 1, 3, 5, 7 or 9).

  8. When counting outcomes across 'either … or …' alternatives, do you add or multiply?

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    Add across alternatives; multiply within each one. E.g. classes: 7×14 + 7×12 + 7×14×12 = 1358.

  9. How many 6-digit codes use each of the digits 1 to 6 exactly once?

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    6 × 5 × 4 × 3 × 2 × 1 = 720.

  10. How many 5-digit even numbers use digits 1, 4, 6, 7, 8 each exactly once?

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    3 × 4 × 3 × 2 × 1 = 72 (fix the last digit as 4, 6 or 8 first, then arrange the rest).

  11. 10 teams each play every other team 6 times. How many matches in total?

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    45 pairs of teams (10 × 9 ÷ 2) × 6 = 270 matches.

  12. How many different pairs can be chosen from 60 girls?

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    60 × 59 ÷ 2 = 1770 (divide by 2 because order within a pair does not matter).

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