GCSE Product rule for counting Practice Questions

Free GCSE Product rule for counting practice questions with full step-by-step worked solutions. Covers product rule, sample space, three stages, counting numbers. Practise exam-style problems and check your method.

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GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
There are 33 tops and 55 pairs of shorts. How many different outfits can be made?
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Worked solution

  1. Each top pairs with each pair of shorts

    3 tops, 5 shorts3\ \text{tops},\ 5\ \text{shorts}

    For each of the 3 tops there are 5 possible shorts.

  2. Product rule: multiply

    3×53\times5

    The product rule says multiply the number of choices at each stage.

  3. Work it out

    3×5=153\times5=15

    So there are 15 outfits.

Answer
1515
Question 2
1 markeasy
A drinks machine has 44 flavours and 22 cup sizes. How many drinks are possible?
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Worked solution

  1. Flavour choices then size choices

    4 flavours, 2 sizes4\ \text{flavours},\ 2\ \text{sizes}

    For each flavour there are 2 cup sizes.

  2. Multiply

    4×24\times2

    Use the product rule.

  3. Work it out

    4×2=84\times2=8

    So there are 8 drinks.

Answer
88
Question 3
2 marksintermediate
A spinner numbered 11-55 is spun three times. How many possible outcomes are there?
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Worked solution

  1. Spinner numbered 1-5, spun 3 times

    5 each5\ \text{each}

    Each spin has 5 outcomes.

  2. First spin

    55

    5 outcomes.

  3. Second spin

    55

    5 outcomes.

  4. Third spin

    55

    5 outcomes.

  5. Product rule

    5×5×5=535\times5\times5=5^3

    Multiply, giving 5³.

  6. Work it out

    53=1255^3=125

    So there are 125 outcomes.

Answer
125125
Question 4
3 markshard
How many four-letter codes can be made from 66 different letters if no letter is repeated?
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Worked solution

  1. Choose 4 of the 6 letters in order

    6 letters6\ \text{letters}

    No letter repeats and order matters.

  2. First position

    66

    6 choices.

  3. Second position

    55

    5 remain.

  4. Third position

    44

    4 remain.

  5. Fourth position

    33

    3 remain.

  6. Product rule

    6×5×4×36\times5\times4\times3

    Multiply the choices.

  7. First multiplication

    6×5=306\times5=30

    30 so far.

  8. Next

    30×4=12030\times4=120

    120 so far.

  9. Last

    120×3=360120\times3=360

    So the total is 360.

  10. State the answer

    360360

    So there are 360 codes.

Answer
360360
Question 5
5 markschallenging
A four-digit PIN uses digits 00-99 with repeats allowed. Use the product rule and the subtraction principle to find how many PINs have at least one repeated digit.
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Worked solution

  1. Count all possible PINs

    10410^4

    4 digits, 0-9, repeats allowed.

  2. Work it out

    104=1000010^4=10000

    So there are 10000 PINs in total.

  3. 'At least one repeat' is hard directly

    count the opposite\text{count the opposite}

    It is easier to count PINs with NO repeat, then subtract.

  4. 'No repeat' means all different

    all digits different\text{all digits different}

    Count PINs where every digit is different.

  5. First digit

    1010

    Any of 10 digits.

  6. Second digit

    99

    9 remain.

  7. Third digit

    88

    8 remain.

  8. Fourth digit

    77

    7 remain.

  9. Count all-different PINs

    10×9×8×710\times9\times8\times7

    Multiply the choices.

  10. Work it out

    10×9×8×7=504010\times9\times8\times7=5040

    So there are 5040 PINs with no repeat.

  11. Use the subtraction principle

    total−no repeat\text{total}-\text{no repeat}

    At least one repeat = total minus no-repeat.

  12. Subtract

    10000−504010000-5040

    Take the no-repeat count away.

  13. Work it out

    10000−5040=496010000-5040=4960

    So there are 4960 PINs with at least one repeat.

  14. Why subtract

    easier opposite\text{easier opposite}

    Counting the no-repeat PINs is far easier than counting the repeats directly.

  15. State the answer

    49604960

    So there are 4960 PINs with at least one repeated digit.

Answer
104−10×9×8×7=10000−5040=496010^4 - 10×9×8×7 = 10000 - 5040 = 4960

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Hard QuestionsWorked Solutions

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