Hard GCSE Product rule for counting Questions

Challenging, exam-style GCSE Product rule for counting questions with worked solutions. Stretch yourself on the hardest product rule, restrictions, no repeats, arrangements problems.

product rulerestrictionsno repeatsarrangementsfactorialsblock method
GCSE Higher34 questionsStep-by-step solutions
Question 1
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.
Show worked solution

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
Question 2
5 markschallenging
Explain why the number of ways to arrange nn different objects in a row is n!n! (n factorial).
Show worked solution

Worked solution

  1. We arrange n different objects in a row

    n objectsn\ \text{objects}

    Fill n positions, one object each.

  2. Fill the positions one at a time

    position by position\text{position by position}

    Think about the choices at each position in turn.

  3. The first position

    n choicesn\ \text{choices}

    Any of the n objects can go first.

  4. One object is now used

    n−1 leftn-1\ \text{left}

    After placing one, n − 1 remain.

  5. The second position

    n−1 choicesn-1\ \text{choices}

    Any of the remaining n − 1 objects.

  6. The third position

    n−2 choicesn-2\ \text{choices}

    Now n − 2 objects remain.

  7. The pattern continues

    n−3, n−4,…n-3,\ n-4,\ldots

    Each new position has one fewer choice.

  8. The last position

    1 choice1\ \text{choice}

    Only one object is left for the final position.

  9. Apply the product rule

    multiply all\text{multiply all}

    Multiply the number of choices at every position.

  10. Write the product

    n×(n−1)×(n−2)×⋯×1n\times(n-1)\times(n-2)\times\cdots\times1

    This is the full product of the choices.

  11. This product has a name

    n!

    It is written n! and read 'n factorial'.

  12. Example n=3n = 3

    3×2×1=63\times2\times1=6

    For 33 objects there are 66 arrangements.

  13. Example n=4n = 4

    4×3×2×1=244\times3\times2\times1=24

    For 44 objects there are 2424 arrangements.

  14. Why it works

    fewer each time\text{fewer each time}

    Each choice uses up an object, reducing the next count by one.

  15. State the conclusion

    n!

    So the number of arrangements of n different objects is n!.

Answer
The first object has nn places, the next n−1n-1, then n−2n-2, ..., down to 11; multiplying gives n×(n−1)×...×1=n!n\times(n-1)\times...\times1 = n!
Question 3
5 markschallenging
On a 3×33\times3 grid you may only move right or up. How many shortest routes are there from the bottom-left to the top-right corner?
Show worked solution

Worked solution

  1. You move right or up

    R or UR\ \text{or}\ U

    Every step is either a Right move or an Up move.

  2. How many of each are needed

    3R, 3U3R,\ 3U

    On a 3×3 grid you must go 3 right and 3 up to reach the far corner.

  3. So each route is a sequence of 66 moves

    6 moves6\ \text{moves}

    Three R's and three U's in some order.

  4. Count the orderings

    arrange RRRUUU\text{arrange } RRRUUU

    The number of routes is the number of arrangements of RRRUUU.

  5. If all 66 were different

    6!6!

    Six distinct items would give 66! orders.

  6. Work out 66!

    6!=7206!=720

    6×5×4×3×2×1=7206 \times 5 \times 4 \times 3 \times 2 \times 1 = 720.

  7. The three R's are identical

    3!3!

    Swapping the R's changes nothing.

  8. The three U's are identical

    3!3!

    Swapping the U's changes nothing.

  9. Divide by both

    6!3!×3!\dfrac{6!}{3!\times3!}

    Remove the repeats among the R's and the U's.

  10. Work out the denominator

    3!×3!=6×6=363!\times3!=6\times6=36

    The bottom is 3636.

  11. Divide

    72036\dfrac{720}{36}

    Divide 720720 by 3636.

  12. Work it out

    72036=20\dfrac{720}{36}=20

    So there are 2020 routes.

  13. Alternative view

    choose 3 of 6 to be R\text{choose 3 of 6 to be }R

    Equivalently, choose which 33 of the 66 moves are Rights, which also gives 2020.

  14. Confirm

    2020

    The count is 2020.

  15. State the answer

    2020

    So there are 2020 shortest routes.

Answer
2020
Question 4
6 markschallenging
Design your own applied counting problem that is solved using the product rule, and give the full solution.
Show worked solution

Worked solution

  1. Aim to design a product-rule problem

    independent stages\text{independent stages}

    The product rule fits problems with several independent choices.

  2. Choose a context

    lunch deal\text{lunch deal}

    A lunch deal with several parts works well.

  3. Stage 11: the sandwich

    4 sandwiches4\ \text{sandwiches}

    Offer 44 sandwiches.

  4. Stage 22: the drink

    3 drinks3\ \text{drinks}

    Offer 33 drinks.

  5. Stage 33: the snack

    2 snacks2\ \text{snacks}

    Offer 22 snacks.

  6. State the question

    how many deals?\text{how many deals?}

    Ask how many different lunch deals are possible.

  7. The choices are independent

    any of each\text{any of each}

    Any sandwich can go with any drink and any snack.

  8. Apply the product rule

    4×3×24\times3\times2

    Multiply the number of choices at each stage.

  9. First multiplication

    4×3=124\times3=12

    1212 so far.

  10. Multiply by 22

    12×2=2412\times2=24

    So there are 2424 deals.

  11. Interpret the answer

    24 deals24\ \text{deals}

    There are 2424 different lunch deals.

  12. Check it fits the rule

    3 stages\text{3 stages}

    Three independent stages, solved by multiplying.

  13. No deal is missed

    all combinations\text{all combinations}

    Every sandwich-drink-snack combination is counted.

  14. No deal repeats

    each distinct\text{each distinct}

    Each combination is different.

  15. State the problem and answer

    2424

    So the designed problem gives 2424 lunch deals.

Answer
e.g. a lunch deal with 44 sandwiches, 33 drinks and 22 snacks: 4×3×2=244\times3\times2 = 24 different deals
Question 5
5 markschallenging
A lock code is 44 digits from 00-99 with repeats allowed. How many codes contain at least one even digit?
Show worked solution

Worked solution

  1. Count all possible codes first

    10410^4

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

  2. Work it out

    104=1000010^4=10000

    So there are 1000010000 codes in total.

  3. 'At least one even' is hard directly

    use the opposite\text{use the opposite}

    Count the codes with NO even digit, then subtract.

  4. 'No even' means all odd

    all digits odd\text{all digits odd}

    Every digit must be odd.

  5. The odd digits

    1,3,5,7,91,3,5,7,9

    There are 55 odd digits.

  6. Each position has 55 odd choices

    5 each5\ \text{each}

    All four positions pick an odd digit.

  7. Count all-odd codes

    5×5×5×55\times5\times5\times5

    Multiply 55 for each position.

  8. As a power

    545^4

    Four 5s5s multiplied.

  9. Work it out

    54=6255^4=625

    So there are 625625 all-odd codes.

  10. Use the subtraction principle

    total−all odd\text{total}-\text{all odd}

    At least one even = total minus all-odd.

  11. Subtract

    10000−62510000-625

    Take the all-odd count away.

  12. Work it out

    10000−625=937510000-625=9375

    So there are 93759375 codes with at least one even digit.

  13. Why subtract

    easier opposite\text{easier opposite}

    Counting the all-odd codes is far easier than counting 'at least one even' directly.

  14. Sense check

    9375<100009375<10000

    Most codes have an even digit, as expected.

  15. State the answer

    93759375

    So there are 93759375 such codes.

Answer
93759375

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