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

In how many different ways can the letters of the word 'GEOGRAPHY' be arranged such that the vowels must always come together?

  1. 2520

  2. 2530

  3. 15130

  4. 15120
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

GEOGRAPHY has 9 letters: G, E, O, G, R, A, P, H, Y. Vowels: E, A, O (3). Consonants: G, G, R, P, H, Y (6). Treating the 3 vowels as one block, we have 7 items to arrange: (EAO), G, G, R, P, H, Y. Arrangements = 7! / 2! (for the two Gs) = 5040 / 2 = 2520. Within the block, the 3 vowels can be arranged in 3! = 6 ways. Total = 2520 * 6 = 15120.

Multiple choice

A clinic assigns each patient a code that is either a single vowel (A, E, I, O, U) or a pair of distinct vowels written in alphabetical order. Each patient must receive a different code. If the clinic must be able to assign codes to at least 10 patients, then what is the least number of vowels out of the given five vowels that must be used?

  1. 1

  2. 2

  3. 3

  4. 4

  5. 5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Number of single vowels = n. Number of pairs = nC2. Total codes = n + nC2 = n + n(n-1)/2 = (2n + n^2 - n) / 2 = (n^2 + n) / 2. We need (n^2 + n) / 2 >= 10, so n^2 + n >= 20. If n=4, (16+4)/2 = 10. So 4 vowels are sufficient.

Multiple choice

A group of 10 students is to be lined up in a straight line. In how many different ways can the students be arranged if two particular students must not stand next to each other?

  1. 2,303,440

  2. 2,309,040

  3. 2,903,040

  4. 3,903,040

  5. 3,902,440
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total arrangements of 10 students is 10!. Arrangements where two specific students are together is 9! * 2!. Subtracting this from total: 10! - 2 * 9! = 9!(10 - 2) = 8 * 9! = 8 * 362,880 = 2,903,040.

Multiple choice

A four-letter code is formed using only the letters A, B, and C, such that both B and C appear at least once in the code. What is the total number of such distinct four-letter codes that can be formed?

  1. 32

  2. 50

  3. 48

  4. 78
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total codes of length 4 using {A, B, C} is 3^4 = 81. We must exclude codes missing B or C. Using inclusion-exclusion: codes missing B = 2^4 = 16, codes missing C = 2^4 = 16, codes missing both (only A) = 1^4 = 1. Total invalid = 16 + 16 - 1 = 31. Valid codes = 81 - 31 = 50.

Multiple choice

How many distinct 7-letter arrangements can be formed from the letters of the word 'INCONCEIVABLE' (without using any letter more times than it appears in the word) such that the letters in the odd positions (1st, 3rd, 5th, 7th) are chosen from the letters that appear exactly once in the word, and the letters in the even positions (2nd, 4th, 6th) are chosen from the letters that appear exactly twice in the word?

  1. 7,260

  2. 7,420

  3. 7,200

  4. 7,380
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

In how many ways can the word VALEDICTORY be arranged so that all vowels come together ?

  1. 8!/4!

  2. 8! × 2!

  3. 8! - 4!

  4. 8! × 4!

  5. None of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

VALEDICTORY has 11 letters: V, A, L, E, D, I, C, T, O, R, Y. Vowels are A, E, I, O (4). Consonants are V, L, D, C, T, R, Y (7). Treating vowels as one unit, we have 7 consonants + 1 unit = 8 items. These can be arranged in 8! ways. The 4 vowels can be arranged in 4! ways. Total = 8! * 4!.

Multiple choice

Select the correct option that indicates the arrangement of the given words in the order in which they appear in an English dictionary 1. Petitionary 2. Petitioning 3. Petition 4. Petitioners 5. Petitioned

  1. 3, 1, 4, 5, 2

  2. 3, 1, 5, 4, 2

  3. 4, 1, 2, 3, 5

  4. 1, 2, 3, 4, 5

  5. 1, 2, 3, 5, 4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Alphabetical order: 3. Petition, 1. Petitionary, 5. Petitioned, 4. Petitioners, 2. Petitioning. Comparing the strings: 'Petition' (3) is shortest. 'Petitiona' (1) comes next. 'Petitione' (5) comes before 'Petitioners' (4). 'Petitioni' (2) comes last.

Multiple choice

In how many distinct ways can the letters of the word 'CRICKET' be rearranged?

  1. 5040

  2. 2520

  3. 720

  4. 1440

  5. 10080
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Box 1: 16 balls total, 2 blue. Box 2: 5 + x balls total, x blue. P(Blue) = 0.5 * (2/16) + 0.5 * (x / (5+x)) >= 0.25. 1/16 + x / (2(5+x)) >= 0.25. x / (2(5+x)) >= 0.25 - 0.0625 = 0.1875. x / (5+x) >= 0.375. x >= 0.375(5+x) = 1.875 + 0.375x. 0.625x >= 1.875. x >= 3.