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Cambridge IGCSE0606

Permutations and combinations

Mathematics Additional 0606 Chapter Notes

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Permutations and combinations
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Permutations and combinations notes

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In this chapter

  1. 1Permutation or combination?
  2. 2Factorial notation
  3. 3Permutations of n items taken r at a time
  4. 4Combinations of n items taken r at a time
  5. 5Arrangements with a restriction
  6. 6Selections with conditions
  7. 7Algebraic problems

1. Permutation or combination?

A permutation is an arrangement: the order of the items matters. A password and a number formed from digits are permutations.

A combination is a selection: the order does not matter. A team, a committee or a group of students is a combination.

Examiner insight

“Many candidates either did not use all of the information in the question or were unable to determine whether order was important.” “Errors in parts (i) and (ii) tended to be from the use of permutations rather than combinations.”

Worked example 14 marks

Five candidates stand for election.

(a) A captain and a vice-captain are chosen.(b) Two of them attend a meeting. For each, say whether it is a permutation or a combination, with a reason, and find the number of ways.

  1. 1

    (a) The two jobs are different, so swapping the two people gives a different outcome: a permutation. [1]

  2. 2

    ⁵P₂ = 5 × 4 = 20 [1]

  3. 3

    (b) Both attend the same meeting, so the order does not matter: a combination. [1]

  4. 4

    ⁵C₂ = (5 × 4)/(2 × 1) = 10 [1]

2. Factorial notation

n! is the product of all the whole numbers from n down to 1. It counts the ways of arranging n different items in a row.

By definition, 0! = 1. Every factorial contains the one before it, so common factors cancel in a fraction.

n! = n(n-1)(n-2) ⋯ 3 × 2 × 1

n! = n × (n-1)!

0! = 1

Examiner insight

“Many candidates were unable to simplify the factorial terms correctly, highlighting a need to practice manipulation of factorials.”

Worked example 13 marks

Find the value of n such that ⁿ⁺³C₆ = 12 × ⁿ⁺²C₅.

  1. 1

    ((n+3)!)/((n-3)! 6!) = 12 × ((n+2)!)/((n-3)! 5!) [1]

  2. 2

    Divide both sides by ((n+2)!)/((n-3)! 5!). The algebraic factorials give ((n+3)!)/((n+2)!) = n+3 and the numerical ones give (5!)/(6!) = 1/6, so (n+3)/6 = 12 [1]

  3. 3

    n + 3 = 72, so n = 69 [1]

3. Permutations of n items taken r at a time

The number of permutations of r items chosen from n different items is ⁿP_r. There are n choices for the first place, n-1 for the second, and so on for r places.

A question asks for a value. Work it out; do not leave ⁹P₆ or 12! as the answer.

ⁿP_r = (n!)/((n-r)!)

ⁿP_r = n(n-1)(n-2) ⋯ (n-r+1)

Examiner insight

“Some candidates obtained 6720. This needed to be multiplied by 6 in order to get the total number of 6-digit numbers.” “A value was expected and 12! was not acceptable as an answer.”

Worked example 13 marks

A 6-digit number is to be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9. No digit may be used more than once in any 6-digit number.

Find how many 6-digit numbers can be formed in the following cases.

(i) There are no further restrictions.

(ii) The 6-digit number includes the digit 4.

  1. 1

    (i) Order matters: ⁹P₆ = 9 × 8 × 7 × 6 × 5 × 4 = 60 480 [1]

  2. 2

    (ii) The digit 4 can go in any of the 6 places: 6 ways [1]

  3. 3

    The other 5 places are filled from the 8 other digits: ⁸P₅ = 6720, so 6 × 6720 = 40 320 [1]

4. Combinations of n items taken r at a time

The number of combinations of r items chosen from n different items is ⁿC_r. Each selection of r items can be arranged in r! ways, so dividing the permutations by r! removes the order. Evaluate the result.

ⁿC_r = (n!)/((n-r)! r!)

ⁿC_r = ⁿP_r/(r!)

Examiner insight

“Most candidates obtained the correct result, but it was essential that 14C6 be evaluated.”

Worked example 14 marks

The number of combinations of r objects out of 20 objects is 4/3 times the number of combinations of r+1 objects out of 20 objects. Find the value of r.

  1. 1

    ²⁰C_r = 4/3 ²⁰C_r+1 [1]

  2. 2

    (20!)/(r! (20-r)!) = 4/3 × (20!)/((r+1)! (20-(r+1))!) [1]

  3. 3

    Cancel 20! and r!, and use (r+1)! = (r+1) r! and (20-r)! = (20-r) (19-r)!: 1/(20-r) = 4/(3(r+1)) [1]

  4. 4

    3r + 3 = 80 - 4r, so 7r = 77 and r = 11 [1]

5. Arrangements with a restriction

Start with the restricted place. A number cannot start with 0. An even number ends in 0, 2, 4, 6 or 8. A multiple of 5 ends in 0 or 5. "Greater than 40 000" limits the first digit.

When the restriction allows more than one case, state each case, work it out, then add the cases. Write the method down: sets of figures with no explanation are meaningless unless they can be identified easily.

Examiner insight

“These candidates needed to realise that a 5-digit number cannot start with a 0.” “In a question of this type, it is essential that the method used is stated clearly to enable examiners to give credit where necessary.”

Worked example 13 marks

A 5-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. No digit may be used more than once in any 5-digit number. Find how many of these 5-digit numbers are greater than 40 000 and divisible by 5.

  1. 1

    Divisible by 5, so the last digit is 0 or 5. Case 1, ends in 0: the first digit is 4 to 9, so 6 ways. The 3 middle digits come from the 8 left: ⁸P₃ = 336. Total 6 × 336 = 2016 [1]

  2. 2

    Case 2, ends in 5: the first digit is 4, 6, 7, 8 or 9, so 5 ways. The 3 middle digits come from the 8 left: ⁸P₃ = 336. Total 5 × 336 = 1680 [1]

  3. 3

    Add the cases: 2016 + 1680 = 3696 [1]

6. Selections with conditions

A condition such as at least 2 girls gives several cases. List them all. Within one case, multiply the combinations (this and that). Across cases, add (this case or that case).

The other method is to subtract from the total every case you do not want. Remove all of them.

Examiner insight

“Some candidates combined their combinations using incorrect operations – commonly adding when they should have been multiplying.” “Some candidates using the subtraction method did not consider all the appropriate cases.”

Worked example 13 marks

A class contains 10 girls and 12 boys. A group of 6 students is to be selected from the class. There must be at least 2 girls and at least 2 boys in the group. Find the number of different selections that can be made.

  1. 1

    Cases: 2 girls and 4 boys, 3 girls and 3 boys, 4 girls and 2 boys [1]

  2. 2

    ¹⁰C₂ × ¹²C₄ + ¹⁰C₃ × ¹²C₃ + ¹⁰C₄ × ¹²C₂ [1]

  3. 3

    = 45 × 495 + 120 × 220 + 210 × 66 = 22 275 + 26 400 + 13 860 = 62 535 [1]

7. Algebraic problems

Write each ⁿC_r or ⁿP_r as factorials. Cancel the common factors using (n+1)! = (n+1) n!, then expand and simplify. If an equation gives a quadratic, discard the solution that cannot be a value of n.

A show that question ends on the result it gives. Keep every line of algebra.

(n+2)! = (n+2)(n+1) n!

ⁿC₃ = (n(n-1)(n-2))/6

Examiner insight

“Most candidates stated the correct equality in terms of factorials, but many went no further.” “A small number of candidates omitted to discard the extraneous solution.”

Worked example 14 marks

Show that, for all values of n geqslant 3, ⁿ⁺²C₃ - ⁿC₃ = n².

  1. 1

    ⁿ⁺²C₃ = ((n+2)!)/((n-1)! 3!) and ⁿC₃ = (n!)/((n-3)! 3!) [1]

  2. 2

    = ((n+2)(n+1)n)/6 - (n(n-1)(n-2))/6 [1]

  3. 3

    = n/6[(n²+3n+2) - (n²-3n+2)] [1]

  4. 4

    = n/6 × 6n = n² [1]

Recap

  • Order matters: permutation. Order does not matter: combination.
  • n! is n(n-1) ⋯ 1, and 0! = 1.
  • ⁿP_r = (n!)/((n-r)!) and ⁿC_r = (n!)/((n-r)! r!). Evaluate the answer.
  • Restricted places first: no leading 0, last digit for even or multiples of 5.
  • Several cases: multiply within a case, add across cases, and list every case.
  • Algebraic questions: write as factorials, cancel, expand, and keep every line.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A coach chooses 3 players from a squad of 8 to take part in a relay. Is the number of ways of choosing the 3 players found using a permutation or a combination?1 mark
  2. Do not use a calculator. Evaluate (7!)/(5!) and state the value of 0!.2 marks
  3. Do not use a calculator. Find the value of ⁹P₃.2 marks
  4. Do not use a calculator. Find the value of ⁸C₃.2 marks
  5. Do not use a calculator. A club has 8 members. A chair, a secretary and a treasurer are chosen from the members. No member holds more than one post. Find the number of ways of filling the three posts.3 marks
  6. A teacher chooses 4 students from 12 to take part in a quiz. Two of the 12 students are twins. Find the number of different groups of 4 that contain both twins.3 marks
  7. Do not use a calculator. A 4-digit even number is formed from the digits 1, 2, 3, 4, 5 and 6. No digit may be used more than once. Find how many such numbers there are.3 marks
  8. Do not use a calculator. A 4-digit odd number is formed from the digits 0, 1, 2, 3, 4, 5 and 6. No digit may be used more than once, and the number does not start with 0. Find how many such numbers there are.4 marks
  9. A group of 6 students is to be chosen for a science fair from 8 Year 10 students and 6 Year 11 students. The group must contain at least 2 students from each year. Two of the Year 10 students are twins, and either both twins are in the group or neither is. Find the number of different groups that can be chosen.6 marks
  10. Do not use a calculator. A chess club has n members. The number of ways of choosing a team of 4 members is 6 times the number of ways of choosing a pair of members. (a) Find the value of n. (b) Find the number of teams of 4 that include the club captain but do not include the club secretary. The captain and the secretary are different members.6 marks

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