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 33 tops there are 55 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 1515 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 22 cup sizes.

  2. Multiply

    4×24\times2

    Use the product rule.

  3. Work it out

    4×2=84\times2=8

    So there are 88 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 1−51-5, spun 33 times

    5 each5\ \text{each}

    Each spin has 55 outcomes.

  2. First spin

    55

    55 outcomes.

  3. Second spin

    55

    55 outcomes.

  4. Third spin

    55

    55 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 125125 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 44 of the 66 letters in order

    6 letters6\ \text{letters}

    No letter repeats and order matters.

  2. First position

    66

    66 choices.

  3. Second position

    55

    55 remain.

  4. Third position

    44

    44 remain.

  5. Fourth position

    33

    33 remain.

  6. Product rule

    6×5×4×36\times5\times4\times3

    Multiply the choices.

  7. First multiplication

    6×5=306\times5=30

    3030 so far.

  8. Next

    30×4=12030\times4=120

    120120 so far.

  9. Last

    120×3=360120\times3=360

    So the total is 360360.

  10. State the answer

    360360

    So there are 360360 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

    44 digits, 0−90-9, repeats allowed.

  2. Work it out

    104=1000010^4=10000

    So there are 1000010000 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 1010 digits.

  6. Second digit

    99

    99 remain.

  7. Third digit

    88

    88 remain.

  8. Fourth digit

    77

    77 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 50405040 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 49604960 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 49604960 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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