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 words can be formed from the letters of the word $'SIGNATURE'$ so that all the vowels come together?
  1. $17280$
  2. $4320$
  3. $720$
  4. $80$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

SIGNATURE has 9 letters: S, G, N, T, R (consonants) and I, A, U, E (vowels). Treat the 4 vowels as one block. We have 5 consonants + 1 block = 6 items. They can be arranged in 6! ways. The 4 vowels can be arranged among themselves in 4! ways. Total = 6! * 4! = 720 * 24 = 17280.

Multiple choice

Consider the word $'PERMUTATION'$. How many permutations can be made such that a word should always start with a vowel?
  1. $\dfrac{5\times 10!}{2}$
  2. $\dfrac{10!}{2!}$
  3. $5!\times 6!$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

PERMUTATION has 11 letters: P, E, R, M, U, T, A, T, I, O, N. Vowels are E, U, A, I, O (5). Consonants are P, R, M, T, T, N (6). Total letters = 11. If it starts with a vowel, there are 5 choices for the first position. The remaining 10 positions can be filled in 10! / 2! ways (due to two T's). Total = 5 * 10! / 2! = 5 * 10! / 2.

Multiple choice

In how many ways can the letters of the word $'FRIDAY'$ be arranged. If the vowels are never together.
  1. $240$
  2. $480$
  3. $120$
  4. $360$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The word FRIDAY has 6 distinct letters, including 2 vowels (I, A) and 4 consonants. The total number of unrestricted arrangements is 6! = 720. If the vowels are kept together as a single unit, there are 5 units to arrange in 5! * 2! = 240 ways. Subtracting this from the total gives 720 - 240 = 480 ways where the vowels are never together.

Multiple choice

There are $13$ letters of $8$ different sorts $I,I,I,S,S,T,T,L,L,A,O,N,D$. In finding groups of $4$, how many permutations can be made if following are the possibilities to be considered? If $2$ are alike of one kind and $2$ are alike of other kind.
  1. $44$
  2. $52$
  3. $36$
  4. $102$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Letters: I,I,I; S,S; T,T; L,L; A; O; N; D. We need 2 alike of one kind and 2 alike of other kind. Pairs available: I, S, T, L (4 pairs). Choose 2 pairs from 4: 4C2 = 6 ways. Permutations for each pair: 4! / (2! * 2!) = 6. Total = 6 * 6 = 36.

Multiple choice

The chance that the vowels are separated in an arrangement of the letters of the word "HORROR" is

  1. 1/2

  2. 2/3

  3. 3/4

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Word 'HORROR' has 6 letters: H(2), O(2), R(2). Total arrangements = 6! / (2!2!2!) = 720 / 8 = 90. Arrangements where vowels (O, O) are together: treat (OO) as one unit, arrangements = 5! / (2!2!) = 120 / 4 = 30. Probability vowels together = 30/90 = 1/3. Probability vowels separated = 1 - 1/3 = 2/3.

Multiple choice

From $6$ different novels and $3$ different dictionaries, $4$ novels and $1$ dictionary are to be selected and arranged in a row on a she. If so that the dictionary is always in the middle. Then the number of such arrangements is
  1. at least $500$ but less than $750$
  2. at least $750$ but less than $1000$
  3. at least $1000$
  4. less than $500$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Select 4 novels from 6: 6C4 = 15. Select 1 dictionary from 3: 3C1 = 3. Arrange 4 novels and 1 dictionary with dictionary in middle: 4! * 1! = 24. Total = 15 * 3 * 24 = 1080.

Multiple choice

A person goes in for an examination in which there are four papers with a maximum of $m$ marks from each paper. The number of ways in which one can get $2 m$ marks is
  1. $ ^{2m+3}C_3 $
  2. $ \dfrac{1}{3} (m+1)(2m^2+4m+1) $
  3. $ \dfrac{1}{3} (m+1)(2m^2+4m+3) $
  4. $ ^{2m-3}C_3 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is a stars and bars problem equivalent to finding the number of non-negative integer solutions to x1 + x2 + x3 + x4 = 2m, where each xi <= m. Using the inclusion-exclusion principle, the total solutions are C(2m+4-1, 4-1) = C(2m+3, 3), and subtracting cases where at least one xi > m gives the result.

Multiple choice

For an examination, a candidate has to select $7$ subjects from $3$ different groups $A,\space B,\space C$ which contain $4,\space 5,\space 6$ subjects, respectively.The number of ways in which a candidate can make his selection if he has to select at least $2$ subjects from each group are
  1. $2700$
  2. $2600$
  3. $2500$
  4. $2400$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total subjects = 7. Groups: A(4), B(5), C(6). Need at least 2 from each. Possible distributions: (2,2,3), (2,3,2), (3,2,2). Ways = (4C2 * 5C2 * 6C3) + (4C2 * 5C3 * 6C2) + (4C3 * 5C2 * 6C2) = (6 * 10 * 20) + (6 * 10 * 15) + (4 * 10 * 15) = 1200 + 900 + 600 = 2700.