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
Answer key: Systematic listing and the product rule for counting worksheet — GCSE Higher
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- (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.
- (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.
- (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.
- (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.
- (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) 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.
- (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.
- (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.
- (d) 720 — Method: the three medals are awarded one after the other, and each award removes one athlete from the pool available for the next, so the product rule multiplies the number of choices at each stage. Working: 10 athletes could take gold; once gold is settled 9 could take silver; once silver is settled 8 could take bronze; so the number of ways is 10 × 9 × 8 = 720. Answer: 720. The distractors: 1000 comes from working out 10 × 10 × 10, which allows the same athlete to take more than one medal; 120 comes from dividing the product by 6, which would be right only if the three medals were identical, whereas gold, silver and bronze are different; 30 comes from multiplying the 10 athletes by the 3 medals instead of multiplying the choices at each stage.
- (b) 6.4 × 10⁸ — There are 5 letter positions, each with 23 choices, and 2 digit positions, each with 10 choices, and every position is independent because repeats are allowed. By the product rule, the total is 23⁵ × 10² = 6,436,343 × 100 = 643,634,300, which is 6.4 × 10⁸ to 2 significant figures. Using all 26 letters instead of the 23 that are actually allowed, ignoring the excluded letters entirely, gives 26⁵ × 10² = 1,188,137,600, which is 1.2 × 10⁹ to 2 significant figures. Adding the seven counts of choices instead of multiplying them, 23 + 23 + 10 + 10 + 23 + 23 + 23, gives 135, which is 1.4 × 10² to 2 significant figures — a total far too small for seven independent positions. Swapping which count of choices belongs to letters and which belongs to digits, working out 23² × 10⁵ instead of 23⁵ × 10², gives 52,900,000, which is 5.3 × 10⁷ to 2 significant figures.
- (c) 11 — Method: work out the total number of combinations as if there were no restriction, then subtract the one combination that is not allowed. Working: without any restriction there are 4 backdrops × 3 outfits = 12 combinations. The grey backdrop with the formal suit is not allowed, removing 1 combination: 12 − 1 = 11. Answer: 11. 12 comes from forgetting to remove the combination that is not allowed. 8 comes from removing the entire formal suit outfit from the count instead of just the one combination with the grey backdrop. 10 comes from removing two combinations instead of just the one that is not allowed.
- (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.
- (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.
- (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) 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.
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