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 general knowledge math & puzzles

The letters of the word ‘EQUATION’ are arranged in a row. Find the probability that all vowels are together

  1. 1/14

  2. 1/21

  3. 1/8

  4. 2/21

  5. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

EQUATION has 8 letters (3 vowels: E,U,A,OI and 5 consonants: Q,T,N). Treating all 4 vowels as one block gives 5! arrangements of blocks, and 4! arrangements within the vowel block. Total with vowels together = 5! × 4! = 120 × 24 = 2880. Total arrangements of 8 letters = 8! = 40320. Probability = 2880/40320 = 1/14.

Multiple choice general knowledge math & puzzles

How many unique 5 letter words can be formed using the letters A,E,I,O,U where all the vowels will be together. repetition of a letter is not allowed.

  1. 96

  2. 48

  3. 120

  4. 0
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since there are 5 vowels (A,E,I,O,U) and they must be together, we treat the block of 5 vowels as a single unit. However, since there are no other letters mentioned to form a word with, we are just arranging the 5 vowels. The number of arrangements is 5! = 120.

Multiple choice general knowledge sports

How many different 9 letter words can be form using the letters of TENDULKAR, so that vowels are always together??

  1. 5040

  2. 8!*3!

  3. 30240

  4. 10!
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

TENDULKAR has 9 letters with 3 vowels (E, U, A) and 6 consonants (T, N, D, L, K, R). Treat the 3 vowels as one block: we arrange 7 items (6 consonants + 1 vowel block) in 7! ways, and arrange the 3 vowels within their block in 3! ways. Total = 7! × 3! = 5040 × 6 = 30240. Option C is correct.

Multiple choice

How many arrangements are possible from the letters (without repetition) of the word COMPUTER, such that vowels are never separated?

  1. 4320

  2. 1540

  3. 720

  4. 1440

  5. 8! - 3!
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

There are three vowels, viz. E, O & U which are not to be separated. Let us assume them to be one letter EOU, then we have got 5 (consonants) + 1(vowels group) = 6 letters which can be arranged in 6 X 5 X 4 X 3 X 2 X 1 = 720 ways. But EOU themselves can be arranged in 3 X 2 X 1 = 6 ways, so total possible arrangements are 720 X 6 = 4320 ways.

Multiple choice

If the letters of the word DAUGHTER are to be arranged so that vowels occupy the odd places, then the number of different words is

  1. 576

  2. 676

  3. 625

  4. 524
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

DAUGHTER has 8 letters with 3 vowels (A, U, E) and 5 consonants (D, G, H, T, R). The 4 odd positions are 1,3,5,7. Arrange 3 vowels in these 4 positions: 4P3 = 4x3x2 = 24 ways. Arrange 5 consonants in remaining 5 positions: 5! = 120 ways. Total = 24 x 120 = 2880. But checking the calculation: 3 vowels in 4 odd places means 4P3 = 24, and 5 consonants in 5 places = 5! = 120, giving 2880 total arrangements. The answer 576 suggests only 3 odd places or different counting.

Multiple choice

The probability that in a random arrangement of letters of the word “MUMBAI”, two M’s do not come together is

  1. 2/3

  2. 1/3

  3. 1/2

  4. 5/7

  5. 14/15
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The word “MUMBAI” has six letters. Therefore, total ways of arrangements = 6!/2 =360. Number of words where two M’s do not come together is = Total words – Number of words where they come together = 360 – 5! = 240 Hence, the required probability is = 240/360 = 2/3.

Multiple choice computer and ms office mathematical methods for economics economics

The number of ways the letter SCHOLAR can be arranged so that LR never come together are .

  1. 5,040

  2. 3,600

  3. 4,404

  4. 3,910
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total arrangements of SCHOLAR (7 letters, all distinct) is 7! = 5040. Arrangements where L and R are together: treat LR as one unit, so 6! * 2! = 720 * 2 = 1440. Subtracting these from total: 5040 - 1440 = 3600.

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

If the letters of the word  $"ATTEMPT"$  are written down at random. The probability that all the  $T's$  come together is

  1. $1/21$
  2. $6/7$
  3. $1/7$
  4. $1/42$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Number of ways of arranging the words keeping all $T's$ together is  $5!$

Number of ways of arranging the words  is  $\dfrac{7!}{3!}$
Probability that all the $T's$  together is $\dfrac{5! }{\dfrac{7!}{3!}}=\dfrac{1}{7}$