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 fanned by jumbling the letters of the word MISSISSIPPI in which no two 's' are adjacent
  1. $\displaystyle 7\cdot ^{6}C_{2}\cdot ^{8}C_{4}$
  2. $\displaystyle 6\cdot 7 \cdot ^{8}C_{4}$
  3. $\displaystyle 6\cdot 8 \cdot ^{7}C_{4}$
  4. $\displaystyle 8\cdot ^{6}C_{4}\cdot ^{7}C_{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

MISSISSIPPI has 4 S's, 4 I's, 2 P's, and 1 M. Arrange the 7 non-S letters in 7!/4!2! ways, then select 4 of the 8 gaps around them for the S's so that no two S's are adjacent. This equals 7 x C(6,2) x C(8,4), matching option A.

Multiple choice

The number of words, that can be formed by using all letters of the word DAUGHTER, if relative order of vowels and consonants remains same, are
  1. $2880$
  2. $6720$
  3. $720$
  4. $3600$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

DAUGHTER has 8 letters: 3 vowels (A, U, E) and 5 consonants (D, G, H, T, R). The relative order of vowels and consonants must remain the same. There are 3! ways to arrange vowels and 5! ways to arrange consonants. Total ways = 3! * 5! = 6 * 120 = 720.

Multiple choice

The letters of the word "LOGARITHM" are arranged in all possible ways. The number of arrangements in which the relative positions of the vowels and consonants are not changed is
  1. $4320$
  2. $720$
  3. $4200$
  4. $3420$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

LOGARITHM has 9 letters: 4 vowels (O, A, I, O is not there, wait: L, O, G, A, R, I, T, H, M). Vowels: O, A, I (3). Consonants: L, G, R, T, H, M (6). Arrangements keeping relative positions: 3! * 6! = 6 * 720 = 4320.

Multiple choice

A researcher plans to identify each participant in a $174$ certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are $12$ participants, and each participant is to receive a different code?
  1. $4$
  2. $5$
  3. $6$
  4. $7$
  5. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

With n letters, the number of available codes is n + n(n - 1)/2. Four letters give only 10 codes, while five letters give 15, so the least number needed is 5.

Multiple choice

A school has 11 Maths teachers and 7 Science teachers. In how many ways can they be seated in a round table so that all the Science teachers do not sit together?
  1. ${ 18! } -( { 12! } \times{ 7! } )  $
  2. ${ 17! } -( { 12! } \times{ 7! } )  $
  3. ${ 17! } -( { 11! } \times{ 7! } )  $
  4. ${ 18! } -( { 11! } \times{ 7! } )  $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total people = 18. Total arrangements in a circle = (18-1)! = 17!. Arrangements where all 7 Science teachers sit together: Treat them as 1 unit. Total units = 11 + 1 = 12. Arrangements = (12-1)! * 7! = 11! * 7!. Total ways = 17! - (11! * 7!).

Multiple choice

How many different words can be formed using all the letters of the word $'ALLAHABAD' $ when all the vowels are together?
  1. $1$
  2. $5040$
  3. $720$
  4. $360$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The word ALLAHABAD has 9 letters: A, A, A, A, L, L, H, B, D. The vowels are A, A, A, A. Treating the four A's as a single block, we arrange the block and the remaining letters (L, L, H, B, D). There are 6 items to arrange: (AAAA), L, L, H, B, D. The number of arrangements is 6! / 2! (for the two L's) = 720 / 2 = 360.

Multiple choice

In how many can the letters of the word $'DETAIL'$ be arranged such that both the vowels and consonants are together
  1. $32$
  2. $36$
  3. $48$
  4. $72$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The word DETAIL has 6 letters: 3 vowels (E, A, I) and 3 consonants (D, T, L). Treating vowels as one block and consonants as one block, we have 2! ways to arrange the blocks. Within the blocks, there are 3! ways to arrange vowels and 3! ways to arrange consonants. Total = 2 * 6 * 6 = 72.