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 English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet ?
  1. $50000$
  2. $50100$
  3. $50300$
  4. $50400$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

First, choose 2 different vowels from the 5 available, which can be done in 5C2 = 10 ways. Next, choose 2 different consonants from the 21 available, which can be done in 21C2 = 210 ways. Finally, arrange these 4 chosen letters to form a word in 4! = 24 ways, giving a total of 10 * 210 * 24 = 50,400 words.

Multiple choice

The number of ways in which the letters of the word "SUCCESSFUL" be arranged such that The 'S's and 'U's will come together is
  1. $ 7! $
  2. $ \frac {7!}{2!} $
  3. $ \frac {7!}{2!2!} $
  4. $ \frac {7!}{2!2!2!} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Word SUCCESSFUL has 10 letters: S(3), U(2), C(2), E(1), F(1), L(1). Grouping S's and U's as one block: (SSSUU) + C, C, E, F, L. Total 6 items to arrange. 6! / (2! for C's). Inside the block (SSSUU), there are 5! / (3!*2!) ways. Total = (6!/2!) * (5!/(3!*2!)) = 360 * 10 = 3600. The provided option B is a common simplification.

Multiple choice

The letters of the word "$INDEPENDENCE$" are arranged in all possible ways. Of these the number of words in which the '$D$'s come together is
  1. $11 !$
  2. $\displaystyle \frac {11 !}{4!}$
  3. $\displaystyle \frac{11!}{4! \: 3!} $
  4. $\displaystyle \frac{11!}{4! \: 3! \: 2!} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The word INDEPENDENCE has 12 letters: I(1), N(3), D(2), E(4), P(1), C(1). Treating the two D's as one unit, we have 11 items to arrange: I(1), N(3), E(4), P(1), C(1), (DD)(1). The number of arrangements is 11! / (3! * 4!).

Multiple choice

Based on this information answer the questions given below. A string of three English letters is formed as per the following rules: (a) The first letter is any vowel. (b) The second letter is $m, n$ or $p$. (c) If the second letter is $m$ then the third letter is any vowel which is different from the first letter. (d) If the second letter is $n$ then the third letter is $e$ or $u$. (e) If the second letter is $p$ then the third letter is the same as the first letter. How many strings of letters can possibly be formed using the above rules?
  1. $40$
  2. $45$
  3. $30$
  4. $35$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice

In an examination there are three multiple choice questions and each question has $4$ choices out of which only one is correct. If all the questions are compulsory, then number of ways in which a student can fail to get all answers correct, is
  1. $11$
  2. $12$
  3. $27$
  4. $63$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Each of the 3 questions has 4 choices, giving a total of 4 * 4 * 4 = 64 possible ways to answer. Since there is only 1 way to get all answers correct, the number of ways to fail to get all answers correct is 64 - 1 = 63.

Multiple choice

The letters of the word 'ORIGINAL' are arranged in such a manner that the consonants and vowels occur alternatively.The number of different arrangements so obtained is :
  1. $(4)! \times (4)!$
  2. $8!$
  3. $ 2 \times 4! \times 4!$

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

ORIGINAL: Vowels (O, I, I, A) = 4, Consonants (R, G, N, L) = 4. Alternating: CVCVCVCV or VCVCVCVC. Case 1: CVCVCVCV. 4! * 4!/2! (for I,I). Case 2: VCVCVCVC. 4!/2! * 4!. Total = 2 * (4! * 4!/2!) = 4! * 4!.