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.
part Higher
Systematic listing and the product rule for counting worksheet — GCSE Higher
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- 1.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.
- 2.Five different books are placed in a row on a shelf. Work out how many different orders the five books can be placed in.
- 3.A board game has 5 different character pieces and 4 different colour tokens. The dragon piece can only be used with the gold token. Work out how many different combinations of one character piece and one colour token are possible.
- 4.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.
- 5.Ten players enter a chess tournament. Every player plays every other player exactly once. Work out how many games are played in the tournament.
- 6.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.
- 7.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.
- 8.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.
- 9.A bag holds 5 different sweets. Noah takes 2 of the sweets out of the bag together, so the order in which he takes them does not matter. Work out how many different pairs of sweets he could take.
- 10.A florist has 5 different flowers. Freya picks 3 of them to make a small bunch, and the order in which she picks them does not matter. Work out how many different bunches she could make.
- 11.A café is planning its lunch menu. It will offer one soup from 3 choices and one sandwich from 4 choices, but the mushroom soup, one of the 3 soups, will only be served with the cheese sandwich, one of the 4 sandwiches, on a normal day. Work out how many different soup-and-sandwich combinations are available on a normal day.
- 12.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.
- 13.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.
- 14.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.
- 15.A padlock code is formed from 3 different digits chosen from 1, 2, 3, 4, 5 and 6 (no digit may be used twice in the same code). Work out how many different codes can be made.
Answer key
- (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.
- (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.
- (d) 17 — Without the restriction there would be 5 × 4 = 20 combinations. The dragon piece can only be paired with the gold token, so of the 4 tokens, 3 are not allowed with the dragon piece, giving 20 − 3 = 17 valid combinations. 20 comes from ignoring the restriction completely. 19 comes from subtracting only 1 of the 3 invalid dragon combinations instead of all 3, 20 − 1 = 19. 16 comes from multiplying only the 4 non-dragon pieces by the 4 tokens, 4 × 4 = 16, and forgetting to add back the one valid combination of the dragon piece with the gold token.
- (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) 45 — Method: count the ordered pairings with the product rule and then correct for the fact that a game between two players is the same game whichever player it is counted from. Working: each of the 10 players meets 9 opponents, so 10 × 9 = 90 pairings are counted; every game has been counted twice, once from each player's side, so the number of games is 90 ÷ 2 = 45. Answer: 45. The distractors: 90 comes from stopping at 10 × 9 and never halving, so that each game is counted once for each of its two players; 55 comes from adding 10 + 9 + 8 + ... + 1 instead of 9 + 8 + ... + 1, which counts one extra round of games; 20 comes from multiplying the 10 players by the 2 players in each game rather than 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.
- (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) 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.
- (d) 10 — Method: list the pairs systematically, taking each sweet in turn and pairing it only with the sweets that come after it, so that no pair is written down twice. Working: numbering the sweets 1 to 5, the first sweet pairs with 4 others, the second pairs with 3 sweets that come after it, the third with 2 and the fourth with 1, so the total is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 20 comes from working out 5 × 4 and never halving, which counts each pair twice, once in each order; 25 comes from working out 5 × 5, which allows the same sweet to be chosen twice; 9 comes from adding the 5 choices and the 4 remaining choices instead of combining them as a selection of two.
- (d) 10 — Method: picking 3 flowers from 5 leaves 2 flowers behind, so counting the different pairs that could be left out counts the bunches, and those pairs can be listed systematically. Working: number the flowers 1 to 5; the first flower can be left out alongside any of the 4 flowers after it, the second alongside any of the 3 after it, the third alongside any of the 2 after it and the fourth alongside the last one, so the number of pairs left out is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 60 comes from working out 5 × 4 × 3 and treating the three picks as an ordered selection when the order does not matter; 30 comes from dividing that product by 2 instead of by the 6 orders in which three chosen flowers could have been picked; 15 comes from multiplying the 5 flowers by the 3 flowers picked instead of counting the selections.
- (d) 9 — Method: split into two cases — the soups with no restriction, and the mushroom soup on its own — then add the totals. Working: the 2 soups other than mushroom can be paired with any of the 4 sandwiches: 2 × 4 = 8. The mushroom soup can only be paired with the cheese sandwich: 1 combination. Total = 8 + 1 = 9. Answer: 9. 12 comes from working out 3 × 4 = 12 without applying the restriction at all. 8 comes from correctly finding the 2 unrestricted soups' 8 combinations, but forgetting to add back the 1 allowed mushroom-and-cheese combination. 11 comes from taking the unrestricted total of 12 and removing only 1 mushroom combination instead of all 3 disallowed ones.
- (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.
- (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.
- (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.
- (a) 120 — There are 6 choices for the first digit. The second digit must be different from the first, leaving 5 choices, and the third digit must differ from both of the first two, leaving 4 choices. By the product rule, the number of codes is 6 × 5 × 4 = 120. Allowing every digit to repeat, ignoring the 'no digit twice' rule entirely, gives 6 × 6 × 6 = 216. Adding the number of choices at each position instead of multiplying them, 6 + 5 + 4, gives 15. Treating the three chosen digits as one unordered set, rather than as digits in a fixed order on the padlock, divides by the 3! = 6 ways of arranging them: 120 ÷ 6 = 20.
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