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Counting Techniques: Permutations and Combinations — Answers

Answer sheet with worked solutions for the Year 11 probability worksheet.

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Success TutoringWorksheetsAnswer sheetYear 11 · Mathematics · Probability

Counting Techniques: Permutations and Combinations — Answers

Part A: The multiplication principle and factorials

Multiplication principle: if one choice can be made in m ways and a second choice in n ways, the two together can be made in m × n ways. Draw a box for each position and write the number of choices in it.

Factorials: n! = n × (n − 1) × … × 2 × 1 is the number of ways to arrange n different objects in a row. By definition 0! = 1.

Worked example. How many 3-digit numbers can be made from the digits 1 to 6 if no digit is repeated? Boxes: 6 × 5 × 4 = 120. If repeats were allowed it would be 6 × 6 × 6 = 216.

1.A PIN is made of 4 digits, each from 0 to 9, and digits may repeat. How many different PINs are possible?

Answer: 10 000

  1. Each of the 4 positions has 10 choices.
  2. 10 × 10 × 10 × 10 = 104 = 10 000
2.In how many different orders can 5 different books be placed on a shelf?

Answer: 120

  1. 5! = 5 × 4 × 3 × 2 × 1 = 120
3.Without a calculator, = (answer space).

Answer: 42

= = 7 × 6 = 42. Cancel the common factorial instead of multiplying everything out.

4.A code is made of 3 different letters chosen from the 26 letters of the alphabet (no letter repeated). How many codes are possible?
  1. A17 576
  2. B15 600
  3. C2600
  4. D78

Answer: 15 600

26 choices for the first letter, 25 for the second and 24 for the third: 26 × 25 × 24 = 15 600. (17 576 = 263 allows repeats; 2600 = 26C3 ignores order.)

Part B: Permutations (order matters)

A permutation is an arrangement where order matters. The number of ways to arrange r objects chosen from n different objects is [sup n]P[sup r] = [frac n!/(n − r)!].

If some objects are identical, divide by the factorial of each repeated group: the letters of a word with n letters, where one letter appears a times and another b times, can be arranged in ways. For objects that must be together, glue them into one block, arrange the blocks, then arrange inside the block.

Worked example. In how many ways can a president and a secretary be chosen from 10 club members? Order matters (the roles differ): 10P2 = 10 × 9 = 90.

5.8 runners compete in a race. In how many ways can the gold, silver and bronze medals be awarded?

Answer: 336

  1. Order matters (gold is different from silver), so use a permutation.
  2. 8P3 = = 8 × 7 × 6 = 336
6.How many different arrangements are there of the letters of the word PARALLEL?

Answer: 3360

  1. PARALLEL has 8 letters: P, A, A, R, L, L, L, E (A appears twice and L three times).
  2. Arrangements = = 40 320 ÷ (2 × 6) = 40 320 ÷ 12 = 3360
7.6 friends sit in a row of 6 seats. In how many ways can they sit if Priya and Sam must sit next to each other?

Answer: 240

  1. Treat Priya and Sam as one block, so there are 5 units to arrange: 5! = 120 ways.
  2. Priya and Sam can swap places within the block: 2! = 2 ways.
  3. Total = 120 × 2 = 240
8.There are 5! = 120 different ways to seat 5 people around a circular table (where only the order around the table matters).

Answer: False

Fix one person's seat to remove the rotations; the other 4 can then be arranged in 4! = 24 ways. So there are (5 − 1)! = 24 arrangements, not 120.

Part C: Combinations (order does not matter)

A combination is a selection where order does not matter. The number of ways to choose r objects from n different objects is [sup n]C[sup r] = [frac n!/r! (n − r)!]. Each combination of r objects can be arranged in r! orders, so nCr = .

Ask yourself: if I swap two of the chosen objects, do I get a different outcome? If yes, use permutations; if no, use combinations. For at least one questions, count everything and subtract the selections with none.

Worked example. How many ways can a team of 3 be chosen from 8 players? 8C3 = = = 56. Compare with 8P3 = 336: each team is counted 3! = 6 times when order matters.

9.Evaluate 10C3.

Answer: 120

  1. 10C3 = = = = 120
10.A committee of 4 is chosen from 7 women and 5 men. How many committees contain exactly 2 women and 2 men?

Answer: 210

  1. Choose the women: 7C2 = 21. Choose the men: 5C2 = 10.
  2. Multiply (both choices are made): 21 × 10 = 210
11.Using the same 7 women and 5 men, how many committees of 4 contain at least one man?

Answer: 460

  1. Use the complement: total committees − committees with no men.
  2. Total: 12C4 = 495. All women: 7C4 = 35.
  3. At least one man: 495 − 35 = 460
12.Which situation should be counted with combinations rather than permutations?
  1. AChoosing 3 students from a class to form a cleaning team
  2. BAwarding first, second and third prizes in an art show
  3. CMaking a 4-digit code with no repeated digits
  4. DArranging 6 photos in a row on a wall

Answer: Choosing 3 students from a class to form a cleaning team

A team of 3 is the same team whatever order the students are picked in, so order does not matter. The other three situations all depend on order.

13.Match each expression to its value.
  • 5P2
  • 5C2
  • 4!
  • 7C1
  • 24
  • 7
  • 20
  • 10

Answer: 5P2 → 20; 5C2 → 10; 4! → 24; 7C1 → 7

Part D: Probability by counting

When every selection or arrangement is equally likely, P(event) = [frac number of favourable selections/total number of selections], and both counts use the same method (both combinations or both permutations).

Worked example. Two cards are chosen at random from 5 cards numbered 1 to 5. Find P(both are even). Total: 5C2 = 10. Favourable: the only even cards are 2 and 4, so 2C2 = 1. P = [frac 1/10].

14.A bag holds 6 red and 4 blue counters. Three counters are drawn at random without replacement. Find the probability that exactly 2 are red.

Answer:

  1. Total selections: 10C3 = 120.
  2. Favourable: choose 2 of the 6 red and 1 of the 4 blue: 6C2 × 4C1 = 15 × 4 = 60.
  3. P(exactly 2 red) = = [frac 1/2]
15.The letters of the word CHANGE are arranged at random in a row. Find the probability that the arrangement begins with C and ends with E.

Answer:

  1. Total arrangements: 6! = 720 (all six letters are different).
  2. Fix C first and E last; the other 4 letters fill the middle in 4! = 24 ways.
  3. P = = [frac 1/30]
16.In a prize draw, 6 numbers are drawn from the numbers 1 to 40. A player chooses 6 numbers. Find the probability that the player matches all 6 numbers drawn.

Answer:

  1. Order does not matter, so the number of possible draws is 40C6 = 3 838 380.
  2. Only one of these matches the player's numbers.
  3. P = [frac 1/3 838 380] (about 2.6 × 10−7)

Part E: Extended problem

A class has 9 girls and 6 boys. A committee of 5 students is chosen at random.

(a) How many different committees are possible?

(b) How many committees contain exactly 3 girls?

(c) Find the probability that the committee contains exactly 3 girls, as a fraction in simplest form and as a decimal to 3 decimal places.

(d) Find the probability that the committee contains at least one boy.

(e) Once chosen, the committee elects a chair and a secretary from its 5 members. In how many ways can these two roles be filled, and why is this a permutation?

17.Answer parts (a) to (e), showing full working.[8 marks]

Model answer: (a) 15C5 = 3003 committees. (b) 9C3 × 6C2 = 84 × 15 = 1260. (c) P = = [frac 60/143] ≈ 0.420. (d) Committees with no boys: 9C5 = 126, so P(at least one boy) = 1 − = = [frac 137/143] ≈ 0.958. (e) 5P2 = 5 × 4 = 20 ways. Order matters because being chair is different from being secretary, so swapping two people gives a different outcome.

Success criteria:

  • (a) uses combinations: 3003 (1 mark)
  • (b) multiplies 9C3 by 6C2 to get 1260 (2 marks)
  • (c) divides by the total and simplifies to ≈ 0.420 (1 mark)
  • (d) uses the complement with 9C5 = 126 to get ≈ 0.958 (2 marks)
  • (e) 20 ways, with a reason that the roles are different so order matters (2 marks)
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