Try these eight questions on paper first. They cover the permutations and combinations chapter in rising difficulty.
Questions
Question 1. From 7 students, (a) choose 3 for a debate team, (b) choose 3 for the roles of speaker, timekeeper and chairperson. Find the number of ways for each.
Answer
(a) Order does not matter: 7C3 = (7 × 6 × 5) ÷ 6 = 35.
(b) Roles differ, so order matters: 7P3 = 7 × 6 × 5 = 210.
Check: 35 × 3! = 210.
Question 2. Evaluate 6P2 and 6C2.
Answer
6P2 = 6 × 5 = 30. 6C2 = 30 ÷ 2! = 15.
Question 3. How many 3-digit odd numbers can be formed from the digits 1, 2, 3, 4 and 5 if no digit is repeated?
Answer
Fill the last digit first: it must be 1, 3 or 5, giving 3 choices.
The first digit then has 4 choices and the second has 3.
The count is 3 × 4 × 3 = 36.
Question 4. Six pupils stand in a row. Find the number of ways in which two named pupils (a) stand together, (b) do not stand together.
Answer
(a) Treat the pair as a block: 5! × 2! = 120 × 2 = 240.
(b) The total is 6! = 720, so the answer is 720 − 240 = 480.
Question 5. A team of 4 is chosen from 6 boys and 5 girls with at least 1 girl. How many teams are there?
Answer
Use the complement. All teams: 11C4 = 330.
Teams with no girl: 6C4 = 15.
The answer is 330 − 15 = 315.
Question 6. How many different arrangements are there of the letters of SUCCESS?
Answer
There are 7 letters: S appears 3 times, C appears 2 times, and U and E appear once each.
The count is 7! ÷ (3! × 2!) = 5 040 ÷ 12 = 420.
Question 7. A student says: “To choose 4 from 10 students, with two friends A and B both included, I compute 10C4 − 2.” Find the correct count and explain the error.
Answer
If A and B are both in, only 2 more members are needed from the other 8: 8C2 = 28.
The student’s method subtracts 2 from all teams, which has no link to the condition. Both n and r must fall by the number of required people.
Question 8. A committee has 2 men and 3 women chosen from 5 men and 6 women. How many committees are possible?
Answer
Choose the men and women separately, then multiply.
5C2 = 10 and 6C3 = 20, so the count is 10 × 20 = 200.
If you got these wrong
- Questions 1 and 2 test deciding whether order matters.
- Questions 3 and 4 test counting arrangements with restrictions.
- Questions 5, 7 and 8 test counting selections with required members.
- Question 6 tests distinguishing over-counting from under-counting.
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