1. LPU MCA Semester 1
  2. PEA515
  3. Unit 5

PEA515 · Unit 5: Permutation, combination and probability

Selection and arrangement problems: numbers, words and geometric applications

30 practice questions · in the ETE syllabus

Key points

  • Permutation if order matters; combination if it does not
  • Repeated letters: divide by factorials of repeats
  • Together: glue and swap; apart: total minus together
  • At least one = total minus none

A permutation is an arrangement in which order matters: P(n, r) = n! / (n - r)!. A combination is a selection in which order does not matter: C(n, r) = n! / (r! (n - r)!).

ProblemCount
Posts filled one by one (president, secretary...)P(n, r)
Committee of r from nC(n, r)
At least one womanAll committees minus those with no woman
Exactly j women and the rest menC(w, j) x C(m, k - j)
Letters of a word with repeatsn! / (p! q! ...) for repeat counts p, q
Two particular people together in a row2 x (n - 1)! (glue them, then swap)
Two particular people not togethern! - 2 x (n - 1)!

Example: the letters of SUCCESS (S three times, C twice) can be arranged in 7! / (3! x 2!) = 420 ways. Selecting 4 from 3 men and 5 women with at least one woman gives C(8, 4) - C(3, 4) = 70 - 0 = 70.

Exam trap: when two people must be together, forgetting that they can swap places halves the answer.

Try two questions

Question 1 · Tier 3, Apply

A committee of 5 is to be formed from 4 men and 5 women. In how many ways can it be formed so that it has at least one woman?

  1. 350
  2. 0
  3. 5
  4. 126
Show the answer

D. All committees C(9, 5) minus those with no woman C(4, 5) = 126.

Why not A: Counts selections with several women more than once.

Why not B: Counts committees with no woman.

Why not C: Counts only committees with exactly one woman.

Question 2 · Tier 2, Understand

A committee of 4 is to be formed from 6 men and 2 women. In how many ways can it be formed so that it has at least one woman?

  1. 70
  2. 15
  3. 40
  4. 55
Show the answer

D. All committees C(8, 4) minus those with no woman C(6, 4) = 55.

Why not A: Ignores the condition.

Why not B: Counts committees with no woman.

Why not C: Counts only committees with exactly one woman.