Permutations and combinations

NUST NET (Engineering) · Maths · Permutations, combinations and probability. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

Both ideas count without listing. A permutation is an arrangement, so order matters: ABC and BAC are different. A combination is a selection, so order does not matter: a team of A, B and C is the same team in any order.

Start every question by asking whether swapping two chosen items gives something new. If yes, use permutations; if no, use combinations. Then use the counting principle: multiply the numbers of ways for steps done one after another (and), add the numbers for separate cases (or).

Conditions are handled by simple tricks. Items that must stay together are tied into one block. Items that must stay apart are placed in the gaps between the others.

Rules to remember

Common mistake

Using permutations for a selection. Choosing a committee or a handshake pair has no order, so use nCr; choosing people for named posts (captain and vice-captain) has order, so use nPr.

Worked example

In how many ways can the letters of the word TUESDAY be arranged so that the three vowels are always together?

  1. The vowels are U, E, A. Tie them into one block: the block and T, S, D, Y make 5 objects.
  2. 5 objects can be arranged in 5! = 120 ways.
  3. Inside the block the 3 vowels can be arranged in 3! = 6 ways.
  4. Total = 120 × 6 = 720.

Answer: 720

Practice questions

Try each one, then open the answer.

1. A committee of 4 is to be chosen from 6 men and 4 women so that it has at least 3 women. The number of ways is

  1. A
    24
  2. B
    28
  3. C
    25
  4. D
    210
Show answer

Answer: C. Exactly 3 women: 4C3 × 6C1 = 24; all 4 women: 1 way. Total 24 + 1 = 25. Choosing 3 women and then any 1 of the other 7 (giving 28) counts some committees twice.

2. How many 3-digit numbers with no repeated digit can be formed from the digits 0, 1, 2, 3, 4?

  1. A
    60
  2. B
    100
  3. C
    48
  4. D
    36
Show answer

Answer: C. The hundreds digit cannot be 0, so it has 4 choices; the tens and units digits then have 4 and 3 choices: 4 × 4 × 3 = 48. 5P3 = 60 wrongly includes arrangements starting with 0.

3. The number of different arrangements of all the letters of the word ENGINEERING is

  1. A
    11!/(3! 3!)
  2. B
    11!/(3! 3! 2!)
  3. C
    11!
  4. D
    11!/(3! 3! 2! 2!)
Show answer

Answer: D. ENGINEERING has 11 letters: E three times, N three times, G twice, I twice and R once. Divide 11! by the factorial of each repeat count.

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