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

The letters ofthe word $\mathrm{C}$O$\mathrm{C}$H$\mathrm{I}$N are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word $\mathrm{C}$O$\mathrm{C}$H$\mathrm{I}$N is
  1. $360$
  2. $192$
  3. $96$
  4. $48$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Word: COCHIN. Letters: C, C, H, I, N, O. Alphabetical: C, C, H, I, N, O. Words starting with C: C C H I N O (1), C C H I O N (1), C C H N I O (1), C C H N O I (1), C C H O I N (1), C C H O N I (1). Total permutations starting with C: 5!/2! = 60. Words before COCHIN: C C ... (24), C H ... (24), C I ... (24), C N ... (24). Total = 96.

Multiple choice

In how many ways can $8$ Indians, $4$ Americans and $4$ Englishmen can be seated in a row so that all the persons of same nationality sit together?
  1. $3!4!8!4!$
  2. $3!8!$
  3. $4!^2$
  4. $8!4!4!$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Treat each nationality group as a single unit. There are 3 groups, which can be arranged in 3! ways. Within each group, the individuals can be arranged in 8!, 4!, and 4! ways respectively. The total number of ways is 3! * 8! * 4! * 4!.

Multiple choice

How many words can be formed by using the letters of the word ' INSURANCE ',when vowels always remain together ?
  1. $8640$
  2. $17280$
  3. $720$
  4. $None$ $of$ $these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The word INSURANCE has 9 letters: I, N, S, U, R, A, N, C, E. Vowels are I, U, A, E (4). Consonants are N, S, R, N, C (5). Treating the 4 vowels as one block, we arrange 6 items (the block + 5 consonants) in 6!/2! ways (due to two Ns). The 4 vowels can be arranged in 4! ways. Total = (6!/2!) * 4! = 360 * 24 = 8640.

Multiple choice

The number of words of four letters containing equal number of vowels and consonants, where repetition is allowed, is
  1. $105^2$
  2. $210\times 243$
  3. $150\times 243$
  4. $150\times 21^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a 4-letter word with equal vowels (V) and consonants (C), we need 2 V and 2 C. There are 5 vowels and 21 consonants. The number of ways to pick positions for 2 vowels is 4C2 = 6. The number of ways to fill these positions is (5^2) * (21^2). Total = 6 * 25 * 441 = 150 * 441 = 150 * 21^2.