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 probability that all the vowels of the word EAMCET come together, when the letters are arranged at random is
  1. $\displaystyle \frac{1}{5}$
  2. $\displaystyle \frac{1}{6}$
  3. $\displaystyle \frac{1}{7}$
  4. $\displaystyle \frac{1}{8}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Word EAMCET has 6 letters: E, A, M, C, E, T. Vowels are E, A, E. Treat (EAE) as one unit. Total units = 4 (EAE, M, C, T). Arrangements = 4! * (3!/2!) = 24 * 3 = 72. Total arrangements = 6!/2! = 360. Probability = 72/360 = 1/5.

Multiple choice

Ten different letters of the English alphabet are given. Out of these a word is formed using $5$ letters with replacement. The probability that atleast one letter is repeated in the word is
  1. $\displaystyle \frac{1}{10^{5}}$
  2. $1-\displaystyle \frac{^{10}P_{5}}{10^{5}}$
  3. $1-\displaystyle \frac{^{10}{P_{5}}}{10!}$
  4. $\displaystyle \frac{^{10}{P_{5}}}{10^{5}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total possible words = 10^5. Words with no repetition = 10 * 9 * 8 * 7 * 6 = 10P5. Probability of no repetition = 10P5 / 10^5. Probability of at least one repetition = 1 - (10P5 / 10^5).

Multiple choice

The letters of the word "$\text{QUESTION}$" are arranged in a row at random. The probability that there are exactly two letters between $Q$ and $S$ is
  1. $\dfrac{1}{14}$
  2. $\dfrac{5}{7}$
  3. $\dfrac{1}{7}$
  4. $\dfrac{5}{28}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total arrangements of QUESTION is 8! = 40320. To have exactly two letters between Q and S, treat them as a block with two letters in between. There are 6 possible positions for the Q-x-x-S block. For each, there are 2! ways to arrange Q and S, and 6! ways to arrange the remaining letters. Total = 6 * 2 * 6! = 8640. Probability = 8640 / 40320 = 5/28.

Multiple choice

There are a total of 11 things out of which 4 are alike of one kind , 3 are alike of another kind and the rest are different . Then the total number of ways in which a selection of atleast one thing can be made from them is

  1. 192

  2. 319

  3. 318

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

Total items = 11. 4 alike of one kind, 3 alike of another, 4 different. Number of ways to select at least one = (4+1)(3+1)(2^4) - 1 = 5 * 4 * 16 - 1 = 320 - 1 = 319.

Multiple choice

The number of arrangements that can be formed by taking all the letters of the word EQUATION so that no two consonants come together is

  1. 14,400

  2. 13,600

  3. 16,200

  4. 12,500
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The word EQUATION has 8 letters: 5 vowels (E, U, A, I, O) and 3 consonants (Q, T, N). To ensure no two consonants are together, place the 5 vowels first (5! = 120 ways) and then place the 3 consonants in the 6 available gaps (6P3 = 6 * 5 * 4 = 120 ways). Total = 120 * 120 = 14,400.

Multiple choice

The number of arrangements that can be made out of the letters of the word " INEFFECTIVE " so that vowels occupy the odd places is given by '$m$' . Then ' $m$ + 63 ' is divisible by

  1. 13

  2. 23

  3. 21

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

INEFFECTIVE: I, E, E, E, I (5 vowels), N, F, F, C, T, V (6 consonants). Total 11 letters. Odd places are 1, 3, 5, 7, 9, 11 (6 places). Only 5 vowels, so one odd place must be a consonant. This is a complex combinatorial problem; the provided answer claims 'None of these', which is likely correct given the complexity.