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

How many different words can be formed using all the letters of the word "PRECIOUSLY" if the letters P, R and E must always remain together?

  1. 2,41,920

  2. 4,02,140

  3. 32,400

  4. 45,320

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Word "PRECIOUSLY" has 10 letters: P, R, E, C, I, O, U, S, L, Y. Treating PRE as one unit, we have 8 units (PRE, C, I, O, U, S, L, Y). These can be arranged in 8! ways. Within the unit PRE, the letters can be arranged in 3! ways. Total = 8! * 3! = 40320 * 6 = 241920.

Multiple choice

There are 4 magazines, 5 novels and 3 plays. In how many ways can you arrange them, such that all the magazines are placed together, all the novels are placed together and all the plays are placed together in the order - magazines, novels and plays.

  1. 17,480

  2. 17,485

  3. 17,280

  4. 17,082

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

The groups (magazines, novels, plays) can be arranged in 1 way (since the order is fixed). Within each group, items can be arranged: 4! for magazines, 5! for novels, 3! for plays. Total ways = 4! * 5! * 3! = 24 * 120 * 6 = 17,280.

Multiple choice

In how many ways can 10 persons (A to J) stand in 2 rows with 5 persons per row, such that A should always be in the second row and C should always be in the first row?

  1. 25 × 8!

  2. 5P2 × 5P2

  3. 4P2 × 4P2

  4. 4! × 4!

  5. 8!
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

There are 10 positions (5 in each row). A must be in row 2 (5 choices), C must be in row 1 (5 choices). The remaining 8 people can be arranged in the remaining 8 spots in 8! ways. Total ways = 5 * 5 * 8! = 25 * 8!.

Multiple choice

In how many different ways can the letters of the word DESIGN be arranged so that the vowels are at the two ends?

  1. 24

  2. 36

  3. 48

  4. 54

  5. 66
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The word DESIGN has 6 letters, consisting of 2 vowels (E, I) and 4 consonants (D, S, G, N). To place the vowels at the two ends, we can arrange them in 2! = 2 ways. The remaining 4 middle positions can be filled by the 4 consonants in 4! = 24 ways, resulting in a total of 2 * 24 = 48 arrangements.

Multiple choice

Two letters from the word 'TERMINATOR' are selected at random. Find the probability that both are vowels.

  1. 1/5

  2. 2/15

  3. 4/15

  4. 7/15
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The word 'TERMINATOR' has 10 letters in total, of which 4 are vowels (E, I, A, and O). The number of ways to choose any 2 letters from 10 is 10C2 = 45, and the number of ways to choose 2 vowels from the 4 available is 4C2 = 6. Thus, the probability that both selected letters are vowels is 6/45, which simplifies to 2/15.

Multiple choice

In a monthly test, the teacher decides that there will be 4 questions, one from Exercises 5, 6, 7 and 8. If there are 10 questions in Exercise 5, 12 in Exercise 6, 14 in Exercise 7 and 9 in Exercise 8, find the number of ways in which four questions may be selected?

  1. 15000

  2. 15120

  3. 15200

  4. 15320
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The number of ways to select one question from each of the four exercises is the product of the number of questions in each exercise: 10 * 12 * 14 * 9 = 15120.