Printable · GCSE Higher · ages 14-16
Systematic listing and the product rule for counting worksheet — GCSE Higher
Fifteen questions on "systematic listing and the product rule for counting" — DfE statement N5. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Systematic listing and the product rule for counting worksheet — GCSE Higher
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- 1.A 3-character PIN starts with one letter chosen from A, B, C, D and E, followed by two different digits chosen from 1 to 9 (no digit may be used twice in the same PIN). Work out how many different PINs are possible.
- 2.A café meal deal is one sandwich, one snack and one drink. There are 4 different sandwiches, 3 different snacks and 2 different drinks to choose from. Work out how many different meal deals are possible.
- 3.A café offers sandwiches with one filling from 5 choices and one type of bread from 4 choices. Cheese and mustard, which is one of the 5 fillings, is not available on gluten-free bread, which is one of the 4 breads. Work out how many different sandwiches are possible.
- 4.Two-digit numbers are made using the digits 1, 2, 3 and 4, and no digit may be used twice in the same number. Work out how many different two-digit numbers can be made.
- 5.Two-digit numbers are formed using the digits 2, 5, 7 and 8, and each digit may be used only once in a number. Work out how many of these two-digit numbers are even.
- 6.A three-digit code is made using the digits 1, 2 and 3, and each digit may be used more than once. Work out how many different three-digit codes can be made.
- 7.A café sells ice cream in three flavours: vanilla, mango and pistachio. Isla buys a cone with two scoops, using two different flavours, one for the bottom scoop and one for the top scoop. Work out how many different cones she could buy.
- 8.A school council must choose a committee of 3 pupils from 8 volunteers. The three places on the committee are all the same, so only which pupils are chosen matters. Work out how many different committees could be formed.
- 9.Three-digit numbers are made using the digits 2, 3, 4, 6, 8 and 9. No digit may be used twice in the same number. Work out how many of these three-digit numbers are odd.
- 10.Two fair six-sided dice are rolled together. One die is red and the other is blue. Work out how many different outcomes are possible.
- 11.A netball team has 8 players. Two of them are chosen to be captains, and the two captains have equal standing. Work out how many different pairs of captains could be chosen.
- 12.By listing systematically, work out how many two-digit multiples of 5 can be made using the digits 0, 3 and 5, if each digit can be used at most once and the number cannot start with 0.
- 13.A restaurant offers a lunch deal of one starter from 5 options, one main from 6 options and one dessert from 3 options. Work out how many different lunch deals are possible.
- 14.Five different books are placed in a row on a shelf. Work out how many different orders the five books can be placed in.
- 15.A car manufacturer offers a car in 6 colours and 4 trim levels. Two colour-and-trim combinations are not available: red with sport trim, and white with sport trim. Work out how many different colour-and-trim combinations are available.
Answer key
- (c) 360 — There are 5 choices for the letter. The first digit can be any of the 9 digits from 1 to 9, giving 9 choices, and the second digit must differ from the first, leaving 8 choices. By the product rule, the number of PINs is 5 × 9 × 8 = 360. Allowing the second digit to repeat the first, ignoring the 'no digit twice' rule, gives 5 × 9 × 9 = 405. Adding the numbers of choices instead of multiplying them, 5 + 9 + 8, gives 22. Treating the pair of digits as an unordered choice, rather than as a first digit followed by a second digit in a fixed order, halves the digit count: 5 × (9 × 8 ÷ 2) = 180.
- (d) 24 — Method: the three parts of the deal are chosen independently, so every sandwich can be taken with every snack and every one of those pairs with every drink; the product rule multiplies the number of choices in each part. Working: there are 4 choices of sandwich and each can be taken with any of the 3 snacks, giving 4 × 3 = 12 sandwich-and-snack pairs; each of those pairs can be completed with either of the 2 drinks, so the number of meal deals is 12 × 2 = 24. Answer: 24. The distractors: 9 comes from adding the choices, 4 + 3 + 2, instead of multiplying them, and a candidate who adds writes that total down as the count; 12 comes from multiplying the sandwiches by the snacks and never bringing the drink into the count at all; 27 comes from adding the items on the menu to get 9 and then multiplying that by the 3 parts of the deal, which counts the menu rather than the combinations.
- (c) 19 — Without any restriction there would be 5 × 4 = 20 different sandwiches. The restriction removes exactly one combination, cheese and mustard on gluten-free bread, so subtract 1: 20 − 1 = 19. 20 comes from ignoring the restriction completely. 15 comes from removing the gluten-free bread altogether, as if none of the fillings were available on it, 5 × 3 = 15. 16 comes from removing the cheese and mustard filling completely, as if it were not available on any bread, 4 × 4 = 16.
- (a) 12 — Method: build the number one place at a time, listing systematically: fix the tens digit, then run through every units digit that is still available. Working: any of the 4 digits can go in the tens place, and once it has been used only 3 digits are left for the units place, so there are 4 × 3 = 12 numbers; listing the numbers that begin with 1 gives 12, 13 and 14, and each of the other three starting digits gives 3 numbers in the same way. Answer: 12. The distractors: 16 comes from working out 4 × 4, which allows a digit to be used twice; 8 comes from multiplying the 4 digits by the 2 places in the number instead of multiplying the choices available at each place; 6 comes from treating a number and its reverse as the same, counting only the unordered pairs of digits.
- (c) 6 — The units digit must be even, so it can be 2 or 8, giving 2 choices. The tens digit can then be any of the remaining 3 digits, since one digit has been used for the units. Multiply: 2 × 3 = 6. 12 comes from working out how many two-digit numbers can be made in total, 4 × 3 = 12, ignoring the requirement that the number is even. 8 comes from choosing the units digit from 2 options and then wrongly allowing any of the 4 digits again for the tens digit, 2 × 4 = 8, which lets a digit repeat. 2 comes from counting only the choices for the units digit and forgetting the tens digit.
- (c) 27 — Each of the 3 digits can be chosen independently for each of the 3 positions, so multiply: 3 × 3 × 3 = 27. 9 comes from multiplying only two of the three positions, 3 × 3, and forgetting the third. 6 comes from working out 3 × 2 × 1 = 6, which counts codes where digits do not repeat, but the question allows repeated digits. 3 comes from considering only one digit position.
- (a) 6 — Method: the two scoops sit in different places on the cone, so a cone is an ordered choice; the possibilities can be listed systematically or counted by multiplying the choices available at each stage. Working: there are 3 flavours for the bottom scoop, and once that flavour is used only 2 flavours remain for the top scoop, so there are 3 × 2 = 6 cones; listing them confirms this, since vanilla on the bottom allows mango or pistachio on top, mango on the bottom allows vanilla or pistachio, and pistachio on the bottom allows vanilla or mango. Answer: 6. The distractors: 3 comes from treating the two scoops as interchangeable, so that vanilla under mango and mango under vanilla are counted as one cone; 9 comes from allowing the same flavour to be used for both scoops, giving 3 × 3; 5 comes from adding the 3 choices for the bottom scoop to the 2 choices left for the top scoop instead of multiplying them.
- (a) 56 — Method: count the ordered selections with the product rule first, then divide by the number of different orders in which any one committee could have been picked. Working: there are 8 choices for a first pupil, 7 for a second and 6 for a third, giving 8 × 7 × 6 = 336 ordered selections; any particular three pupils could have been picked in 3 × 2 × 1 = 6 orders, so the number of different committees is 336 ÷ 6 = 56. Answer: 56. The distractors: 336 comes from stopping at 8 × 7 × 6 and treating the three places as distinct posts when they are identical; 168 comes from dividing that product by 2 rather than by the 6 orders in which three chosen pupils can be listed; 24 comes from multiplying the 8 volunteers by the 3 places instead of multiplying the choices at each stage.
- (a) 40 — Method: a number is odd exactly when its units digit is odd, so the restricted position is filled first and the two free positions are then filled from the digits that are left, multiplying the number of choices at each stage. Working: of the six digits only 3 and 9 are odd, so there are 2 choices for the units digit; once that digit has been used, 5 digits remain for the hundreds position and then 4 remain for the tens position, so the count is 2 × 5 × 4 = 40. Answer: 40. The distractors: 120 comes from ignoring the word odd altogether and counting every three-digit number that can be made from the six digits, 6 × 5 × 4; 60 comes from filling the hundreds and tens positions first, 6 then 5, and only then allowing 2 odd digits for the units position, which overcounts because one of 3 and 9 may already have been used, giving 6 × 5 × 2; 72 comes from restricting the units digit to 3 or 9 correctly but overlooking the condition that no digit may be used twice, so all six digits are still counted as available for each of the other two positions, giving 2 × 6 × 6.
- (c) 36 — Method: set the outcomes out in a grid with one die along the top and the other down the side, so that every cell of the grid is one outcome, and count the cells. Working: the red die can land in 6 ways, so the grid has 6 columns, and the blue die can also land in 6 ways, so the grid has 6 rows; the number of cells is 6 × 6 = 36. Answer: 36. The distractors: 12 comes from adding 6 and 6 instead of multiplying them; 6 comes from counting the outcomes of a single die and forgetting that the second die also has to land; 21 comes from treating the two dice as indistinguishable, so that a red 2 with a blue 3 and a red 3 with a blue 2 are counted as one outcome.
- (b) 28 — Method: count the ordered choices with the product rule and then correct for the double counting, because the two captains have equal standing and so a pair is the same pair whichever captain is named first. Working: there are 8 players who could be named first and 7 who could be named second, giving 8 × 7 = 56 ordered choices; each pair has been counted twice, once in each order, so the number of pairs is 56 ÷ 2 = 28. Answer: 28. The distractors: 56 comes from stopping at 8 × 7 and never halving, which counts each pair of captains twice; 64 comes from working out 8 × 8, which allows the same player to be chosen as both captains; 16 comes from multiplying the 8 players by the 2 captaincies instead of pairing the players with one another.
- (b) 3 — Method: list all valid two-digit numbers that can be made without starting with 0, then keep only the ones that are multiples of 5. Working: the two-digit numbers possible are 30, 35, 50 and 53. A number is a multiple of 5 only if it ends in 0 or 5: 30 ends in 0, 35 ends in 5, 50 ends in 0, but 53 ends in 3. So there are 3 multiples of 5. Answer: 3. 4 comes from including 53 as a multiple of 5 without checking that its last digit is not 0 or 5. 2 comes from leaving out 50, wrongly assuming 0 cannot be used as the second digit either. 6 comes from listing every two-digit arrangement of the three digits, including ones that start with 0, without applying either restriction.
- (c) 90 — Multiply the number of choices for each course: 5 × 6 × 3 = 90. 14 comes from adding the three numbers instead of multiplying them. 30 comes from multiplying only the starters and mains, 5 × 6, and forgetting the dessert. 18 comes from multiplying only the mains and desserts, 6 × 3, and forgetting the starter.
- (b) 120 — Method: fill the positions on the shelf one at a time; each book placed leaves one fewer book available for the next position, and the product rule multiplies the choices. Working: there are 5 books for the first position, 4 for the second, 3 for the third, 2 for the fourth and 1 for the last, so the number of orders is 5 × 4 × 3 × 2 × 1 = 120. Answer: 120. The distractors: 25 comes from multiplying the 5 books by the 5 positions rather than multiplying the shrinking number of choices at each position; 60 comes from halving the correct product, as though each order had been counted twice in the way that pairs are; 720 comes from carrying the product one factor too far and working out 6 × 5 × 4 × 3 × 2 × 1, as though there were six books.
- (c) 22 — Without restriction there are 6 × 4 = 24 combinations. Two specific combinations are not available, so subtract 2: 24 − 2 = 22. 24 comes from ignoring the restriction completely. 23 comes from subtracting only 1 of the 2 excluded combinations. 18 comes from removing the whole sport trim level, 6 × 3 = 18, instead of removing just the two excluded combinations.
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