Quantitative Aptitude

Permutation and Combination

609 Questions
All Vowels or Objects Together 139 Restricted Group Selection 135 Objects Never Together 98 Fundamental Counting Principle 96 Objects with Repetitions Arrangement 72 Rank or Dictionary Word 20 Selection With Identical Items 15 Pairing or Grouping 10 Distributing Distinct Objects 9 Distributing Identical Objects 9 Circular Ring Arrangement 5 Circular Necklace Arrangement 1

Permutation and Combination Questions

Multiple choice

In the word "EXPERIMENT," how many different words, not essentially meaningful, can be formed using all letters, if each word must begin with the letter "T"?

  1. 60,448

  2. 60,480

  3. 60,840

  4. 64,840
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The word EXPERIMENT has 10 letters: E(3), X(1), P(1), R(1), I(1), M(1), N(1), T(1). Fixing T at the start, we arrange the remaining 9 letters (E, X, P, E, R, I, M, E, N). The number of arrangements is 9! / 3! = 362880 / 6 = 60480.

Multiple choice

In how many distinct ways can the letters of the word "TRIANGLE" be arranged such that the vowels are placed only in positions that are prime numbers?

  1. 2,880

  2. 880

  3. 280

  4. 80
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The word TRIANGLE has 8 letters: 3 vowels (I, A, E) and 5 consonants (T, R, N, G, L). Prime positions are 2, 3, 5, 7. There are 4 prime positions for 3 vowels. Ways to choose and arrange vowels = 4P3 = 4*3*2 = 24. Ways to arrange 5 consonants in remaining 5 positions = 5! = 120. Total ways = 24 * 120 = 2880.

Multiple choice

The letters of the word EQUATION are arranged in such a way that all vowels as well as consonants are together. How many such arrangements are there?

  1. 240

  2. 720

  3. 1440

  4. 1620
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The word EQUATION has 8 distinct letters: 5 vowels (E, U, A, I, O) and 3 consonants (Q, T, N). Treating vowels as one block and consonants as another, we have 2! ways to arrange the blocks, 5! ways to arrange vowels within their block, and 3! ways to arrange consonants within their block. The total is 2 * 120 * 6 = 1440.

Multiple choice

In how many ways can the letters of the word GARGANTUNG be rearranged such that all the G’s appear together?

  1. 8!/ 3!2!2!

  2. 8!/2!2!

  3. 8! .3!/2!2!

  4. 8!/2!.3!

  5. 10!/3!.2!.2!
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

GARGANTUNG has 10 letters: G(3), A(2), R(1), N(2), T(1), U(1). Treating the three G's as one block, we arrange 8 items (the block + 7 remaining letters). The remaining letters are A(2), R(1), N(2), T(1), U(1). Total arrangements = 8! / (2! * 2!).

Multiple choice

How many different four-letter words can be formed (the words need not be meaningful) using the letters of the word GREGARIOUS such that each word starts with G and ends with R?

  1. 8P2

  2. 8P2/2!.2!

  3. 8P4

  4. 8P4/2!.2!

  5. 10P2/2!.2!
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The word GREGARIOUS has 10 letters: G, R, E, G, A, R, I, O, U, S. We need a 4-letter word starting with G and ending with R. The middle two letters must be chosen from the remaining 8 letters (R, E, G, A, I, O, U, S). Since we need to pick 2 letters from 8 and arrange them, it is 8P2.

Multiple choice

A, B, C, D, E, I are sitting in a row. 1. D has only 1 adjacent person. 2. No two people with vowel names sit together. 3. I has only 1 adjacent person, and that is to his left. How many seating arrangements are possible?

  1. 4

  2. 1

  3. 2

  4. 6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

People: A, E, I (vowels); B, C, D (consonants). 6 positions. I is at the far left (pos 1). D is at the far right (pos 6). Remaining: A, E, B, C. Vowels cannot be adjacent. Possible arrangements for A, E, B, C in pos 2, 3, 4, 5: B, A, C, E; B, E, C, A; C, A, B, E; C, E, B, A. Total 4.