Permutation and Combination: Additional Mathematics questions

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

7! / 5! = ______

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7!/5! = 7 × 6 = 42. 35 uses 7 × 5; 210 is 7P3.

For the teacher: Remembering, low difficulty, standard 4.1.2

Textbook p. 122

A cafe offers 3 starters, 4 main dishes and 2 drinks. A meal has one of each. How many different meals are possible?

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Multiply the choices: 3 × 4 × 2 = 24. 9 comes from adding.

For the teacher: Understanding, low difficulty, standard 4.1.1

Textbook p. 121

How many ways can the 5 different letters of the word STUDY be arranged in a row?

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5! = 120. 24 is 4!, and 720 is 6!.

For the teacher: Remembering, low difficulty, standard 4.1.2

Textbook p. 122

6 people sit around a round table. In how many different ways can they sit?

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Circular permutations: (6 − 1)! = 5! = 120. 720 = 6! ignores that rotating the circle gives the same arrangement.

For the teacher: Understanding, low difficulty, standard 4.1.2

Textbook p. 123

How many different arrangements can be made from all the letters of the word BANANA?

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6 letters, A three times, N twice: 6!/(3! × 2!) = 720/12 = 60.

For the teacher: Applying, low difficulty, standard 4.1.2

Textbook p. 127

5 different books are arranged in a row on a shelf. In how many ways can this be done if 2 particular books must be together?

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Treat the pair as one block: 4! = 24 arrangements, and the pair swaps in 2! = 2 ways. 24 × 2 = 48.

For the teacher: Applying, medium difficulty, standard 4.1.2

Textbook p. 128

How many even 4-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if no digit is repeated?

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Last digit: 3 even choices. Other three places: 5P3 = 60. Total 3 × 60 = 180. 360 ignores the even condition.

For the teacher: Applying, medium difficulty, standard 4.1.3

Textbook p. 130

How many ways can 3 pupils be chosen from 10 pupils to attend a meeting?

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Order does not matter: 10C3 = (10 × 9 × 8)/(3 × 2 × 1) = 120. 720 is 10P3.

For the teacher: Applying, low difficulty, standard 4.2.2

Textbook p. 133

A group has 5 boys and 6 girls. How many ways can 2 boys and 3 girls be selected?

200
30
150
2400

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5C2 × 6C3 = 10 × 20 = 200. 30 comes from adding 10 + 20; 2400 uses permutations.

For the teacher: Applying, high difficulty, standard 4.2.3

Textbook p. 135

Which situation needs a permutation rather than a combination?

Choosing 3 members for a committee from 8 people
Choosing 2 fruits from 5 kinds for a salad
Awarding gold, silver and bronze medals to 3 of 8 runners
Selecting 4 books from 12 to borrow

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The three medals are different, so the order of the runners matters. The other situations only select a group.

For the teacher: Analysing, low difficulty, standard 4.2.1

Textbook p. 132

Given 12Cr = 12C4 and r ≠ 4, find the value of r.

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nCr = nC(n − r), so r = 12 − 4 = 8.

For the teacher: Understanding, medium difficulty, standard 4.2.2

Textbook p. 133

Given nP2 = 56, find the value of n.

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nP2 = n(n − 1) = 56. 8 × 7 = 56, so n = 8.

For the teacher: Applying, medium difficulty, standard 4.1.2

Textbook p. 125

Section B: structured questions

33 marks

Answer all questions. Show your working.

Aina has 4 blouses, 3 skirts and 2 scarves.

[2 marks] (a)Find the number of different outfits of one blouse, one skirt and one scarf.

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4 × 3 × 2 [1]1
= 24 [1]1

Tick each point you got.

[1 mark] (b)Find the number of outfits without a scarf.

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4 × 3 = 12 [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 4.1.1

Textbook p. 120

Calculate each value.

[1 mark] (a)6!

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720 [1]1

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[1 mark] (b)9! / 7!

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9 × 8 = 72 [1]1

Tick each point you got.

[1 mark] (c)8P3

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8 × 7 × 6 = 336 [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 4.1.2

Textbook p. 122

The letters of the word MALAYSIA are arranged in a row.

[2 marks] (a)Find the number of different arrangements.

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8 letters, A three times: 8!/3! [1]1
= 6720 [1]1

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[2 marks] (b)Find the number of arrangements in which the three letters A are together.

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Block AAA with M, L, Y, S, I: 6 objects [1]1
6! = 720 [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 4.1.2

Textbook p. 127

3 boys and 2 girls stand in a row for a photograph.

[1 mark] (a)Find the number of ways they can stand with no condition.

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5! = 120 [1]1

Tick each point you got.

[2 marks] (b)Find the number of ways in which the 2 girls are next to each other.

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Block of girls + 3 boys = 4 objects: 4! = 24 [1]1
24 × 2! = 48 [1]1

Tick each point you got.

[1 mark] (c)Find the number of ways in which the 2 girls are not next to each other.

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120 − 48 = 72 [1]1

Tick each point you got.

For the teacher: Analysing, low difficulty, standard 4.1.3

Textbook p. 128

A 4-digit number is formed from the digits 0, 1, 2, 3, 4, 5. No digit is repeated.

[2 marks] (a)Find the number of 4-digit numbers that can be formed.

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First digit not 0: 5 ways [1]1
5 × 5 × 4 × 3 = 300 [1]1

Tick each point you got.

[3 marks] (b)Find the number of even numbers.

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Last digit 0: 5 × 4 × 3 = 60 [1]1
Last digit 2 or 4: 2 × 4 × 4 × 3 = 96 [1]1
60 + 96 = 156 [1]1

Tick each point you got.

[2 marks] (c)Find the number of numbers greater than 4000.

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First digit 4 or 5: 2 ways [1]1
2 × 5 × 4 × 3 = 120 [1]1

Tick each point you got.

For the teacher: Applying, high difficulty, standard 4.1.3

Textbook p. 129

A team of 4 is chosen from 5 boys and 4 girls.

[1 mark] (a)Find the number of ways with no condition.

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9C4 = 126 [1]1

Tick each point you got.

[2 marks] (b)Find the number of ways for a team of 2 boys and 2 girls.

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5C2 × 4C2 [1]1
10 × 6 = 60 [1]1

Tick each point you got.

[2 marks] (c)Find the number of ways for a team with at least 1 girl.

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No girl: 5C4 = 5 [1]1
126 − 5 = 121 [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 4.2.3

Textbook p. 135

Amir says: "The number of ways to choose a president and a secretary from 7 members is 7C2 = 21."

[2 marks] (a)Is Amir correct? Explain.

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No [1]1
The two posts are different so order matters; permutation: 7P2 = 42 [1]1

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[1 mark] (b)Find the number of ways to choose 2 representatives from the 7 members.

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7C2 = 21 [1]1

Tick each point you got.

For the teacher: Evaluating, medium difficulty, standard 4.2.1

Textbook p. 132

7 different people sit around a round table.

[1 mark] (a)Find the number of ways they can sit.

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(7 − 1)! = 720 [1]1

Tick each point you got.

[2 marks] (b)Find the number of ways in which 2 particular people sit next to each other.

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Block + 5 others: (6 − 1)! = 120 [1]1
120 × 2! = 240 [1]1

Tick each point you got.

[1 mark] (c)7 different beads are threaded on a bracelet, so clockwise and anticlockwise arrangements are the same. Find the number of different bracelets.

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720 ÷ 2 = 360 [1]1

Tick each point you got.

For the teacher: Analysing, medium difficulty, standard 4.1.2

Textbook p. 125

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Section B: structured questions

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20