Target Exam

CUET

Subject

General Aptitude Test

Chapter

Quantitative Reasoning

Topic

Permutation & Combination

Question:

Match List-I with List-II

List-I

List-II

(A) The number of different words that can be formed with CUSTOM with the condition that the word should begin with M is _____.

(I) 70

(B) The number of different ways in which the letters of the word EXTRA can be arranged so that the vowels are never together is _____.

(II) 45

(C) There are 10 points in a plane. No three of these points are in a straight line. The total number of straight line that can be formed by joining the two points is ______.

(III) 72

(D) The number of ways a committee of 4 people be chosen out of 8 people is _____.

(IV) 120

Choose the correct answer from the options given below:

Options:

(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

(A)-(II), (B)-(I), (C)-(III), (D)-(IV)

(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Correct Answer:

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Explanation:

The correct answer is Option (4) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

List-I

List-II

(A) The number of different words that can be formed with CUSTOM with the condition that the word should begin with M is _____.

(IV) 120

(B) The number of different ways in which the letters of the word EXTRA can be arranged so that the vowels are never together is _____.

(III) 72

(C) There are 10 points in a plane. No three of these points are in a straight line. The total number of straight line that can be formed by joining the two points is ______.

(II) 45

(D) The number of ways a committee of 4 people be chosen out of 8 people is _____.

(I) 70

(A) Words from "CUSTOM" beginning with 'M'

The word CUSTOM has 6 distinct letters: C, U, S, T, O, M.

  • If the word must begin with M, we fix 'M' in the first position: M _ _ _ _ _
  • The remaining 5 positions can be filled by the remaining 5 letters (C, U, S, T, O) in $5!$ ways.
  • Calculation: $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
  • Match: (A) $\rightarrow$ (IV)

(B) "EXTRA" arrangements where vowels are never together

The word EXTRA has 5 distinct letters: E, X, T, R, A.

  • Total arrangements: $5! = 120$.
  • Arrangements with vowels together: Treat the vowels (E, A) as a single unit. We now have 4 units: (EA), X, T, R.
    • Ways to arrange these 4 units: $4! = 24$.
    • Internal arrangements of the vowels within the unit: $2! = 2$.
    • Total "together" cases: $24 \times 2 = 48$.
  • Vowels never together: $\text{Total} - \text{Together} = 120 - 48 = 72$.
  • Match: (B) $\rightarrow$ (III)

(C) Straight lines from 10 points

A straight line is formed by joining any 2 points. If no three points are in a straight line (non-collinear), the number of lines is the number of ways to choose 2 points out of 10.

  • Formula: ${}^{10}C_2 = \frac{10 \times 9}{2 \times 1} = 45$.
  • Match: (C) $\rightarrow$ (II)

(D) Committee of 4 out of 8 people

The number of ways to choose 4 people out of 8 is a combination problem.

  • Formula: ${}^8C_4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.
  • Match: (D) $\rightarrow$ (I)