These eleven questions cover the whole of permutations and combinations, ordered from easier to harder. All questions are original, and no calculator is needed if you cancel carefully. Check the Cambridge Additional Mathematics 0606 syllabus guidance for your exam year on calculator use.
How to use it: write one line on whether order matters before you calculate. Attempt each question on paper, then open the answer and compare both the method and the number. Log any slip in the mistake log and retest queue, and use the non-calculator working trainer if the cancelling is slow.
Questions
1. Evaluate 7P3 and 7C3.
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7P3 = 7 × 6 × 5 = 210. 7C3 = 210 / 3! = 210 / 6 = 35.
Check: 35 × 6 = 210, so nPr = nCr × r! holds.
2. A class of 10 students needs (a) a prefect and a deputy prefect, and (b) two students to attend a workshop. Find the number of ways for each.
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(a) The two roles are different, so order matters: 10P2 = 10 × 9 = 90.
(b) The two students have the same role, so order does not matter: 10C2 = 90 / 2 = 45.
Check: 45 × 2 = 90.
3. Six different books are placed on a shelf. Find the number of arrangements (a) with no condition, and (b) with the Mathematics book at the left end.
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(a) 6! = 720.
(b) The Mathematics book is fixed, so the other 5 books are arranged freely: 5! = 120.
Check: the Mathematics book could be at any of 6 positions equally, and 720 / 6 = 120.
4. How many 3-digit odd numbers can be formed from the digits 1 to 6, with no digit repeated?
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Fill the last digit first. It must be odd: 1, 3 or 5, so 3 choices. Then 5 choices for the first digit and 4 for the second.
3 × 5 × 4 = 60.
Check: the total number of 3-digit numbers is 6 × 5 × 4 = 120, and half of them are odd, so 60.
5. A committee of 5 is chosen from 7 men and 6 women. How many committees have exactly 3 women?
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3 women and 2 men: 6C3 × 7C2 = 20 × 21 = 420.
Check: 6C3 = (6 × 5 × 4) / 6 = 20. 7C2 = (7 × 6) / 2 = 21. 20 × 21 = 420.
6. A team of 4 is chosen from 10 players, and the captain, Dev, must be in the team. How many teams are possible?
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Place Dev, then choose 3 more from the other 9: 9C3 = (9 × 8 × 7) / 6 = 504 / 6 = 84.
Check: each of the 10C4 = 210 teams has 4 members, so Dev is in 210 × 4 / 10 = 84 of them.
7. Six people stand in a row. In how many arrangements are A and B next to each other?
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Treat A and B as one block. The 5 objects can be arranged in 5! = 120 ways, and A and B can swap inside the block in 2 ways.
120 × 2 = 240.
Check: A and B are next to each other in 5 possible pairs of positions, and there are 2 orders, so 10 ways to place them. The other 4 people fill the remaining places in 4! = 24 ways. 10 × 24 = 240.
8. A group of 4 is chosen from 5 boys and 7 girls. How many groups contain at least one boy?
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Use the complement. Total: 12C4 = (12 × 11 × 10 × 9) / 24 = 11 880 / 24 = 495. No boys, so all 4 from 7 girls: 7C4 = 35.
495 − 35 = 460.
Check (direct): 1 boy 5 × 35 = 175. 2 boys 10 × 21 = 210. 3 boys 10 × 7 = 70. 4 boys 5. Sum = 175 + 210 + 70 + 5 = 460.
9. Seven people stand in a row. In how many ways are P and Q not next to each other?
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Total 7! = 5040. P and Q adjacent: 6! × 2 = 720 × 2 = 1440.
5040 − 1440 = 3600.
Check (gaps): arrange the other 5 in 5! = 120 ways, which leaves 6 gaps. P and Q go in two different gaps in 6 × 5 = 30 ways. 120 × 30 = 3600.
10. A group of 4 is chosen from 5 boys and 4 girls. A student says the number of groups with at least one girl is 4 × 8C3 = 224, by choosing one girl first and then any 3 of the other 8 people. Explain the error, then find the correct answer.
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Error: a group with two girls is counted twice (once for each girl chosen first), with three girls three times, and with four girls four times.
Check of 224: 1 girl 4 × 10 = 40, counted once = 40. 2 girls 6 × 10 = 60, counted twice = 120. 3 girls 4 × 5 = 20, counted three times = 60. 4 girls 1, counted four times = 4. The sum is 40 + 120 + 60 + 4 = 224, which confirms the source of the error.
Correct answer: 9C4 − 5C4 = 126 − 5 = 121.
Check: 40 + 60 + 20 + 1 = 121.
11. How many 4-digit even numbers can be formed from the digits 0 to 9 with no digit repeated and no leading zero?
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Split by the last digit, because 0 behaves differently.
Last digit 0: the other three places use the 9 non-zero digits: 9 × 8 × 7 = 504.
Last digit 2, 4, 6 or 8 (4 choices): the first digit cannot be 0 or the last digit, so 8 choices. The second digit has 8 choices (zero is allowed now, and two digits are used). The third has 7. So 4 × 8 × 8 × 7 = 1792.
Total = 504 + 1792 = 2296.
Check (complement): numbers with no leading zero and no repeats: 9 × 9 × 8 × 7 = 4536. Odd ones: last digit has 5 choices, the first 8, the second 8, the third 7, so 5 × 8 × 8 × 7 = 2240. Then 4536 − 2240 = 2296.
If you got these wrong
Match your slip to the lesson that repairs it.
| What went wrong | Questions | Go to |
|---|---|---|
| Used nPr for a group, or nCr for an ordered job | 1, 2 | Decide whether order matters |
| Condition on a position was applied too late | 3, 4, 11 | Count arrangements with a fixed position |
| Added instead of multiplied, or missed a case | 5, 6, 8 | Count selections with a restriction |
| Long case lists, adjacent and not adjacent | 7, 8, 9 | Use a complement to simplify counting |
| A method sounded right but gave too large a number | 10 | Explain overcounting in a proposed method |
Return to the module overview if you want the study order again. A teacher in online one-to-one Additional Mathematics tuition can go through your attempts and help you choose which lesson to revisit first.