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 how many different ways can the letters of the word \"MACHINE\" be arranged such that the vowels occupy only the odd positions?

  1. 144

  2. 5040

  3. 348

  4. 576

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

MACHINE has 7 letters: 3 vowels (A, I, E) and 4 consonants (M, C, H, N). Vowels must occupy odd positions (1, 3, 5, 7). There are 4 odd positions and 3 vowels, so 4P3 = 24 ways. The 4 consonants occupy the remaining 4 positions in 4! = 24 ways. Total = 24 * 24 = 576.

Multiple choice

Directions: Study the given information and answer the following question. In an examination, there are 3 sections, each containing 5 questions. In the 1st section, the right answer carries 10 marks and the wrong answer fetches 3 negative marks. Second and third sections carry 8 marks and 6 marks per correct answer, respectively and the negative markings per wrong answer are 2 and 1 for 2nd and 3rd sections, respectively. If a question is not attempted, it results in '0' mark. A student has done 11 questions right with all the questions right from exactly one section. If he does not get the same number of right questions in any two sections, then the number of different combinations of attempting right questions is

  1. 3

  2. 6

  3. 12

  4. 14

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

In how many different ways can the numbers 3576124 be arranged using each digit only once in each arrangement such that the digits 6 and 7 are at the extreme ends in each arrangement?

  1. 120

  2. 240

  3. 720

  4. 5040

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

Digits 6 and 7 must be at extreme ends, with 2! arrangements (6 at left, 7 at right OR 7 at left, 6 at right). Remaining 5 digits (3, 5, 7, 1, 2, 4 but wait, 7 is used at ends, so remaining are 3, 5, 1, 2, 4) can be arranged in middle 5 positions in 5! = 120 ways. Total = 2! × 5! = 2 × 120 = 240. This matches option B.

Multiple choice

In how many different ways can the letters of the word 'THOUSANDS' be arranged?

  1. 60480

  2. 120960

  3. 20160

  4. 720

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

The word 'THOUSANDS' has 9 distinct letters. The number of arrangements is 9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362,880. This value is not among options A (60,480), B (120,960), C (20,160), or D (720). Therefore, option E (None of these) is correct.