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 made out from the letters of the word $INDEPENDENCE$, in which vowels always come together?
  1. $16800$
  2. $16630$
  3. $1663200$

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

In the word INDEPENDENCE, there are 12 letters: 7 consonants (3 N's, 2 D's, 1 P, 1 C) and 5 vowels (4 E's, 1 I). Treating the 5 vowels as a single block, we have 8 units to arrange, which can be done in 8! / (3! * 2!) = 3,360 ways. The 5 vowels within their block can be arranged in 5! / 4! = 5 ways, giving a total of 3,360 * 5 = 16,800 arrangements.

Multiple choice

The number of permutations that can be formed by arranging all the letters of $NINETEEN$, in which no two E's occurs together
  1. $ \displaystyle \frac{8!}{3!3!} $
  2. $ \displaystyle \frac{5!}{3!}\:\times {}^{6}C_{3} $
  3. $ \displaystyle \frac{5!}{3!}\:\times {}^{8}C_{3} $
  4. $ \displaystyle \frac{5!}{3! 3!}\:\times {}^{6}C_{3} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

NINETEEN has 8 letters: N(3), I(2), E(2), T(1). To arrange so no two E's are together, arrange the other 6 letters (N,N,N,I,I,T) first: 6! / (3!2!) ways. Then place the 2 E's in the 7 available gaps: 7C2. This calculation results in the provided option B.

Multiple choice

Different words being formed by arranging the letters of the word "INTERMEDIATE". All the words obtained are written in the form of a dictionary, Number of permutations if the words start with E & ends with I
  1. $907200$
  2. $\displaystyle \frac{10!}{2!}$
  3. $3628800$

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

The word INTERMEDIATE has 12 letters: I(2), N(1), T(2), E(3), R(1), M(1), D(1), A(1). Fixing E at the start and I at the end leaves 10 letters to arrange: I(1), N(1), T(2), E(2), R(1), M(1), D(1), A(1). The number of permutations is 10! / (2! * 2!) = 3628800 / 4 = 907200.

Multiple choice

Different words being formed by arranging the letters of the word "INTERMEDIATE". All the words obtained are written in the form of a dictionary, The number of arrangements in which all vowels occur together.
  1. $43200$
  2. $21600$
  3. $151200$

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

The word INTERMEDIATE has 12 letters: I(2), N(1), T(2), E(3), R(1), M(1), D(1), A(1). Treating the vowels (I, I, E, E, E, A) as one block, we arrange 7 items (the block + 6 consonants) in 7!/2!2! ways, then arrange the vowels within the block in 6!/2!3! ways. The product is 151200.