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 exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

In an examination hall, there are four rows of chairs. Each row has $8$ chairs one behind the other. There are two classes sitting for the examination with $16$ students in each class. It is desired that in each row all students belong to the same class and that no two adjacent rows are allotted to the same class. In how many ways can these $32$ students be seated?

  1. $2 \times 16! \times 16!$
  2. $4 \times 16! \times 16!$
  3. $2 \times 8! \times 16!$

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

Since there are 4 rows, let us label the rows as row 1,2,3,4.

each row has 8 chairs. since all the students of the same class sit in the same row. and no adjacent row is alloted to the same class.

therefore one class can be alloted either in 1 and 3 rows or 2 and 4 rows. therefore there are 2 ways to allot the rows to the class.

now 16 students of this class can be arranged in 16 seats, the number of ways to arrange 16 students in 16 seats=16!

similarly 16 students of other class can be arranged in 16! ways.

therefore total number of ways=2$\times$16!$\times$16! ways

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

Find the number of ways in which the letters of the word $'AEROPLANE'$ can be arranged such that the vowels are always together.

  1. $\dfrac{5!}{2!}$
  2. $\dfrac{5!}{2!^2}$
  3. $\dfrac{5!^2}{2!^2}$
  4. $\dfrac{9!}{2!^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$'AEROPLANE'$ Total number of letters $=9$
Vowels - $ 2A,2E,1O$
Considering all vowels as a single letter, we are left with $5$ letters
They can be arranged in $5!$ ways 
Now vowels can be arranged among themselves in $\dfrac{5!}{2!.2!}$
So required number of ways $=\dfrac{5!^2}{2!^2} = 3600 $

Multiple choice

Find the number of ways to arrange the letters in the word "MATHEMATICS" such that the vowels (A, E, I, O, U) always come together.

  1. 1260

  2. 2520

  3. 3780

  4. 5040
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The vowels in the word "MATHEMATICS" are AEIOU. We can consider them as a single unit. Then, we have 8 letters (including the unit of vowels) to arrange. The number of ways to arrange 8 letters is 8!. However, the vowels can be arranged among themselves in 5! ways. Therefore, the number of ways to arrange the letters in the word "MATHEMATICS" such that the vowels always come together is 8! / 5! = 3780.

Multiple choice

In a class of 30 students, how many ways can a teacher select 5 students to participate in a debate competition?

  1. 142506

  2. 2598960

  3. 1454928

  4. 15504
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this problem, we can use the binomial coefficient (\binom{n}{k}). In this case, we have n = 30 (total number of students) and k = 5 (number of students to be selected). So, the number of ways to select 5 students from a class of 30 students is (\binom{30}{5} = \frac{30!}{5!25!} = \frac{30\cdot29\cdot28\cdot27\cdot26}{5\cdot4\cdot3\cdot2\cdot1} = 142506).

Multiple choice

A survey asks respondents to select their favorite color from a list of 5 colors. How many different ways can the respondents answer the question?

  1. 5

  2. 10

  3. 15

  4. 20
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Each respondent can choose only one color, so the number of different ways to answer the question is equal to the number of colors, which is 5.

Multiple choice

In a group of 10 people, how many different ways can you select a committee of 4 people?

  1. 210

  2. 252

  3. 300

  4. 240
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a classic combinatorics problem involving combinations. The formula for combinations is C(n, r) = n! / (n-r)!, where n is the total number of items and r is the number of items to be selected. In this case, n = 10 and r = 4, so C(10, 4) = 10! / (10-4)! = 10! / 6! = 210.

Multiple choice

In a group of 15 people, how many ways can you select a president, a vice president, and a secretary if each person can only hold one position?

  1. 2730

  2. 455

  3. 2205

  4. 3465
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is a classic combinatorics problem involving permutations. The formula for permutations is P(n, r) = n! / (n-r)!, where n is the total number of items and r is the number of items to be selected. In this case, n = 15 and r = 3, so P(15, 3) = 15! / (15-3)! = 15! / 12! = 2205.

Multiple choice

In an objective type examination, 120 objective type questions are there; each with 4 options P, Q, R and S. A candidate can choose either one of these options or can leave the question unanswered. How many different ways exist for answering this question paper?

  1. 5120

  2. 4120

  3. 1205

  4. 1204
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Each of the 120 questions has 5 possible outcomes (P, Q, R, S, or unanswered). Since each question is independent, the total number of ways is 5 * 5 * ... * 5 (120 times), which is 5^120.

Multiple choice

The number of words that can be formed out of the letters of the word ARTICLE so that the vowels occupy even places is :

  1. 574

  2. 36

  3. 754

  4. 144
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

ARTICLE has 7 letters: A, R, T, I, C, L, E. Vowels are A, I, E (3). Consonants are R, T, C, L (4). Even positions are 2, 4, 6. We place the 3 vowels in these 3 positions in 3! = 6 ways. We place the 4 consonants in the remaining 4 positions in 4! = 24 ways. Total = 6 * 24 = 144.

Multiple choice

If the letters of the word (\"MISSISSIPPI\") are written down at random, in a row the probability that no two 'S' occur together is:

  1. 5/33

  2. 7/33

  3. 6/31

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

Total letters = 11 (M:1, I:4, S:4, P:2). Total arrangements = 11!/(4!4!2!). Arrangements with no two S together: place other 7 letters (M,I,I,I,I,P,P) in 7!/(4!2!) ways. There are 8 gaps to place 4 S's: 8C4. Probability = (8C4 * 7!/(4!2!)) / (11!/(4!4!2!)) = 7/33.

Multiple choice

Four-letter words are formed using 17 consonants and 5 vowels. How many will have 2 different vowels in the middle and a consonant at each end?

  1. 5780

  2. 5440

  3. 6700

  4. 7225
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To form the four-letter word with different letters, we need to place two different vowels in the middle and two different consonants at the ends. The number of ways to choose and arrange 2 different vowels from 5 is 5 * 4 = 20. The number of ways to choose and arrange 2 different consonants from 17 for the ends is 17 * 16 = 272. Multiplying these gives 20 * 272 = 5440 words.