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

Directions : Select the correct alternative from the given choices. A, B, C, D, E, F, G and H are 8 students in a school. The class teacher wants to send 4 students each for a quiz competition and drawing competition such that each student goes for exactly one competition. Assuming that each student is capable of participating in either of the two competitions, in how many ways can the class teacher create the two teams?

  1. 70

  2. 72

  3. 75

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

We need to choose 4 students out of 8 for the first competition; the remaining 4 automatically go to the second. The number of ways is 8C4 = (8*7*6*5) / (4*3*2*1) = 70.

Multiple choice

In how many ways can the letters of the word MANAGEMENT be arranged such that no two vowels appear together?

  1. 37,800

  2. 75,600

  3. 25,200

  4. 21,600
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

MANAGEMENT has 10 letters: M, A, N, A, G, E, M, E, N, T. Vowels: A, A, E, E (4). Consonants: M, N, G, M, N, T (6). Arrange consonants: 6! / (2! * 2!) = 720 / 4 = 180. Spaces for vowels: _ C _ C _ C _ C _ C _ C _ (7 spaces). Choose 4 spaces: 7C4 = 35. Arrange vowels: 4! / (2! * 2!) = 6. Total = 180 * 35 * 6 = 37800.

Multiple choice

In how many ways can 13 different alphabets (a, b, c.......m) be arranged so that the alphabets f and g never come together?

  1. 13! - 12!

  2. 13! - 12!/2!

  3. 13! - 2 x 12!

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total arrangements of 13 letters is 13!. Treating 'fg' as a single block, there are 12! ways to arrange the items, and 2! ways to arrange f and g within the block. Thus, the number of ways they are together is 2 * 12!. Subtracting this from total gives 13! - 2 * 12!.

Multiple choice

Consider the word 'QUESTION': How many 8-letter words, with or without meaning, can be formed such that consonants and vowels occupy alternate positions?

  1. 288

  2. 576

  3. 1,152

  4. 2,304
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

QUESTION has 8 letters: Q, U, E, S, T, I, O, N. Consonants: Q, S, T, N (4). Vowels: U, E, I, O (4). To alternate, we can have CVCVCVCV or VCVCVCVC. For CVCVCVCV, 4! ways for consonants and 4! for vowels = 24 * 24 = 576. For VCVCVCVC, 4! * 4! = 576. Total = 576 + 576 = 1152.

Multiple choice

Consider the word 'QUESTION': How many 8-letter words, with or without meaning, can be formed so that all consonants are together?

  1. 5,760

  2. 2,880

  3. 1,440

  4. 720
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

QUESTION has 8 letters: Q, U, E, S, T, I, O, N. Consonants are Q, S, T, N (4). Vowels are U, E, I, O (4). Treating consonants as one block, we have 5 units (block + 4 vowels). These can be arranged in 5! ways. The 4 consonants within the block can be arranged in 4! ways. Total = 120 * 24 = 2880.

Multiple choice

An organizer plans to identify each volunteer at a health-awareness fair with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the minimum number of letters needed if there are 18 volunteers, and each volunteer must receive a different code?

  1. 5

  2. 6

  3. 7

  4. 8

  5. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Codes are single letters (n) or pairs (nC2). Total codes = n + n(n-1)/2 = (2n + n^2 - n)/2 = (n^2 + n)/2. We need (n^2 + n)/2 >= 18. n^2 + n >= 36. If n=5, 25+5=30 < 36. If n=6, 36+6=42 > 36. So 6 letters are needed.