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

Direction: Meena, Nikhil, Orchid, Peter, Quazim and Ralph are sitting around a round table studying for their final examination which is only two days away. Quazim and Nikhil want to sit next to one another because they both have to give the psychology exam and want to discuss. Mina doesn`t like to sit next to Nikhil and she wants to share her Mathematics notes with Peter. Peter prefers to sit next to Orchid because she is the most intelligent person of the group. If Ralph sits to the right of Orchid, how many different arrangements are possible?

  1. 1

  2. 2

  3. 3

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

This is a circular arrangement problem. With constraints (Quazim-Nikhil together, Meena not next to Nikhil, Meena next to Peter, Peter next to Orchid, Ralph right of Orchid), the number of valid arrangements is limited. Given the constraints, there is only 1 valid arrangement.

Multiple choice

In how many different ways the letters of the word 'SHAKESPEAR' can be arranged such that no two 'A' come together. Repetition of letters is not allowed

  1. 10!

  2. 9!

  3. 9!/2!

  4. 10!/2!
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The word SHAKESPEAR has 10 letters with two A's. To ensure no two A's are together, we arrange the 8 non-A letters in 8! ways, then place the 2 A's in the 9 available gaps (including ends) using 9P2 ways. The total is 8! * (9 * 8 / 2) * 2 = 9!.

Multiple choice

Directions: Answer the question independently. Two words USURY and LUXURY are joined together. Words of four letters (all different) are formed with the help of the letters of the new word thus formed. How many such new words can be formed (not necessarily meaningful)?

  1. 360

  2. 330

  3. 17325

  4. 3600
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

USURY has 5 letters (U, S, U, R, Y). LUXURY has 6 letters (L, U, X, U, R, Y). Combined: U, S, U, R, Y, L, U, X, U, R, Y. Unique letters: U, S, R, Y, L, X (6 total). We need to form 4-letter words from these 6 distinct letters. Permutations = 6P4 = 6*5*4*3 = 360.

Multiple choice

Directions: Answer the question independently. In how many different ways can 5 identical apples, 5 identical cherries and 10 identical oranges be arranged such that no two oranges are adjacent to each other?

  1. 2757

  2. 2745

  3. 2772

  4. 792
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Arrange the 5 apples and 5 cherries first (10 items total). The number of ways to arrange these is 10! / (5! * 5!) = 252. There are 11 possible slots for the 10 identical oranges. We must choose 10 slots out of 11, which is C(11, 10) = 11. Total ways = 252 * 11 = 2772.

Multiple choice

Directions: Answer the questions independently of each other. How many different words (not necessarily meaningful) can be formed if all the letters of the word 'RAVINA' are to be arranged such that 'I' is always somewhere between the 2 A's?

  1. 24

  2. 120

  3. 64

  4. 60
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total letters = 6 (R, A, V, I, N, A). Total arrangements = 6! / 2! = 360. The condition 'I is between the two A's' means in any arrangement of the 3 letters (A, I, A), there is only 1 valid order (A, I, A). Since there are 3 positions for these 3 letters out of 6, we divide the total arrangements by 3 (the number of ways to arrange A, I, A). 360 / 3 = 120.

Multiple choice

DIRECTIONS: Type in your answer in the input box provided below the question. In how many ways can the letters of the word NATION be arranged such that no two vowels occupy adjacent positions?

  1. 72

  2. 70

  3. 68

  4. 60
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

NATION has 6 letters: N, A, T, I, O, N. Vowels are A, I, O (3). Consonants are N, T, N (3). Arrange consonants first: 3!/2! = 3 ways (N, T, N). There are 4 gaps between consonants: _ C _ C _ C _. Place 3 vowels in 4 gaps: 4P3 = 4*3*2 = 24. Total = 3 * 24 = 72.