Permutation and Combination MCQ Quiz - Objective Question with Answer for Permutation and Combination - Download Free PDF

Last updated on Jul 22, 2025

An all-inclusive set of Permutation and Combination MCQs Quiz to help you practise for the Quantitative Aptitude section of Banking & Insurance exams as well as high school and graduate exams. Attempting these sets of Permutation and Combination Objective Questions will help you clear the entrance and interview rounds with high scores. These Permutation and Combination Question and Answers have been designed for maximum candidate support and will provide you the necessary training to ace the exam.

Latest Permutation and Combination MCQ Objective Questions

Permutation and Combination Question 1:

How many different ways a 5 members group can be formed out of 4 men and 5 women, the group have at least 2 men?

  1. 105
  2. 162
  3. 221
  4. 262
  5. 321

Answer (Detailed Solution Below)

Option 1 : 105

Permutation and Combination Question 1 Detailed Solution

Given:

Total men = 4

Total women = 5

Group size = 5

Condition: At least 2 men in the group

Formula used:

Number of ways =

Calculations:

Possible combinations with at least 2 men:

Case 1: 2 men, 3 women ⇒ = 6 × 10 = 60

Case 2: 3 men, 2 women ⇒ = 4 × 10 = 40

Case 3: 4 men, 1 woman ⇒ = 1 × 5 = 5

Total ways = 60 + 40 + 5 = 105

∴ The group can be formed in 105 different ways.

Permutation and Combination Question 2:

There are 8 boys and 7 girls out of which a team of 5 players to be selected. In how many ways the team can be selected if at least 2 girls and 2 boys should be in the team?

  1. 2456
  2. 2156
  3. 2486
  4. 2336
  5. 2786

Answer (Detailed Solution Below)

Option 2 : 2156

Permutation and Combination Question 2 Detailed Solution

Calculation

Boys = 8, Girls = 7
Select 5 with at least 2 boys & 2 girls

Valid combinations:

2B + 3G = 8C2 × 7C3 = 28 × 35 = 980

3B + 2G = 8C3 × 7C2 = 56 × 21 = 1176

Total = 980 + 1176 = 2156

Permutation and Combination Question 3:

In how many ways can the letters of the word ‘APPLE' be arranged?

  1. 720
  2. 120
  3. 60
  4. 180

Answer (Detailed Solution Below)

Option 3 : 60

Permutation and Combination Question 3 Detailed Solution

Given:

The word is 'APPLE'.

Formula used:

Number of arrangements of letters = n! / (p! × q! × ...), where n is the total number of letters and p, q, ... are the factorial of the frequency of repeated letters.

Calculation:

Total letters = 5 (A, P, P, L, E)

Repeated letters: 'P' appears 2 times.

⇒ Number of arrangements = 5! / 2!

⇒ 5! = 5 × 4 × 3 × 2 × 1 = 120

⇒ 2! = 2 × 1 = 2

⇒ Number of arrangements = 120 / 2 = 60

∴ The correct answer is option (3).

Permutation and Combination Question 4:

A shop has 4 distinct flavors of ice-cream. One can purchase any number of scoops of any flavor. The order in which the scoops are purchased is inconsequential. If one wants to purchase 3 scoops of ice-cream, in how many ways can one make that purchase? 

  1. 4
  2. 20
  3. 24
  4. 48

Answer (Detailed Solution Below)

Option 2 : 20

Permutation and Combination Question 4 Detailed Solution

CONCEPT:

Combination with Repetition

  • The problem of purchasing 3 scoops of ice-cream from 4 distinct flavors, where the order of selection does not matter, is a classic example of "combinations with repetition."
  • In combinations with repetition, we use the formula:

    C(n + r - 1, r) = (n + r - 1)! / [r! × (n - 1)!]

    • n = number of distinct items (flavors),
    • r = number of selections (scoops).

EXPLANATION:

  • In the given problem:
    • n = 4 (distinct flavors)
    • r = 3 (scoops of ice-cream)
  • Using the formula:
    • C(4 + 3 - 1, 3) = C(6, 3)
    • C(6, 3) = 6! / [3! × (6 - 3)!]
    • C(6, 3) = 6 × 5 × 4 / (3 × 2 × 1)
    • C(6, 3) = 120 / 6
    • C(6, 3) = 20
  • Thus, there are 20 ways to purchase 3 scoops of ice-cream from 4 distinct flavors.

Therefore, the correct answer is Option 2: 20.

Permutation and Combination Question 5:

Fifteen distinct points are randomly placed on the circumference of a circle. At most how many triangles can be formed using these points?

  1. 105
  2. 455
  3. 2730
  4. 30

Answer (Detailed Solution Below)

Option 2 : 455

Permutation and Combination Question 5 Detailed Solution

Given:

Number of distinct points = 15

Formula used:

Maximum number of triangles = C(n, 3)

Where, C(n, r) = 

Calculation:

Maximum number of triangles = C(15, 3)

C(15, 3) = 

C(15, 3) = 

⇒ C(15, 3) = 455

∴ The correct answer is option (2).

Top Permutation and Combination MCQ Objective Questions

How many possible two-digit numbers can be formed by using the digits 3, 5 and 7 (repetition of digits is allowed)?

  1. 10
  2. 9
  3. 7
  4. 8

Answer (Detailed Solution Below)

Option 2 : 9

Permutation and Combination Question 6 Detailed Solution

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⇒ Number of possible two-digit numbers which can be formed by using the digits 3, 5 and 7 = 3 × 3.

∴ 9 possible two-digit numbers can be formed.

The 9 possible two-digit numbers are:

33, 35, 37, 53, 55, 57, 73, 75, 77 

In a meeting of 45 people, there are 40 people who know one another and the remaining know no one. People who know each other only hug, whereas those who do not know each other only shake hands. How many handshakes occur in this meeting?

  1. 225
  2. 10
  3. 210
  4. 200

Answer (Detailed Solution Below)

Option 3 : 210

Permutation and Combination Question 7 Detailed Solution

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Explanation:

Total people in meeting is 45 and out of then 40 people know each other.

So 5 people don't know anyone.

Let those 5 peoples be A, B, C, D, E

So A will handshake 44 people.

B will handshake 43 people

will handshake 42 people

D will handshake 41 people

and E will handshake 40 people

Hence total handshake = 44 + 43 + 42 + 41 + 40 = 210

Option (3) is correct

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

Answer (Detailed Solution Below)

Option 4 : 15120

Permutation and Combination Question 8 Detailed Solution

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Given:

The given number is 'GEOGRAPHY'

Calculation:

The word 'GEOGRAPHY' has 9 letters. It has the vowels E, O, A in it, and these 3 vowels must always come together. Hence these 3 vowels can be grouped and considered as a single letter. That is, GGRPHY(EOA).

Let 7 letters in this word but in these 7 letters, 'G' occurs 2 times, but the rest of the letters are different.

Now,

The number of ways to arrange these letters = 7!/2!

⇒ 7 × 6 × 5 × 4 × 3 = 2520

In the 3 vowels(EOA), all vowels are different

The number of ways to arrange these vowels = 3!

⇒ 3 × 2 × 1 = 6

Now, 

The required number of ways = 2520 × 6 

⇒ 15120

∴ The required number of ways is 15120.

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that atleast 3 men are there on the committee. In how many ways can it be done ?

  1. 645
  2. 564
  3. 735
  4. 756

Answer (Detailed Solution Below)

Option 4 : 756

Permutation and Combination Question 9 Detailed Solution

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Given: 

(7 men + 6 women) 5 persons are to be chosen for a committee.

Formula used:

nCr = n!/(n - r)! r!

Calculation:

Ways in which at least 3 men are selected;

⇒ 3 men + 2 women

⇒ 4 men + 1 woman 

⇒ 5 men + 0 woman 

Number of ways = 7C3 × 6C2 + 7C4 × 6C1 + 7C5 × 6C0

⇒ 7!/(3! × 4!) × 6!/(2! × 4!) + 7!/(4! × 3!) × 6!/(1! × 5!) + 7!/(5! × 2!) × 6!/(6!× 0!)

⇒ 35 × 15 + 35 × 6 + 21 

⇒ 735 + 21 = 756

∴ The required no of ways = 756.

Important PointsThe value of 0! is 1.

Find the number of ways in which 448 mobile phones can be shared equally among students.

  1. 14
  2. 12
  3. 16
  4. 18

Answer (Detailed Solution Below)

Option 1 : 14

Permutation and Combination Question 10 Detailed Solution

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448 = 2 × 2 × 2 × 2 × 2 × 2 × 7

⇒ 448 = 26 × 71

∴ The required number of mobile phones be shared equally in the students = (6 + 1) × (1 + 1) = 7 × 2 = 14

In how many different ways can the letters of the word 'FIGHT' be arranged?

  1. 110
  2. 120
  3. 105
  4. 115

Answer (Detailed Solution Below)

Option 2 : 120

Permutation and Combination Question 11 Detailed Solution

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Given 

Total alphabets in word 'FIGHT' = 5 

Concept Used 

Total number ways of arrangement = n! 

Calculation 

The number of different ways of arrangement of n different words (without repetition) = 5! 

⇒ 5 × 4 × 3 × 2 × 1 = 120 

∴ The required answer is 120 

How many 3 digit odd numbers can be formed from the digits 5, 6, 7, 8, 9, if the digits can be repeated

  1. 55
  2. 75
  3. 70
  4. 85

Answer (Detailed Solution Below)

Option 2 : 75

Permutation and Combination Question 12 Detailed Solution

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Given:

5, 6, 7, 8, 9 are the digits to form 3 digit number

Calculation:

Let us take the 3digit number as H T U (Hundreds, tens, unit digit)  respectively

To make 3 digit number as odd

5, 7, 9 are only possibly be used in the unit digit place

In hundreds and tens place all  5 digits are possible 

Number of ways for unit digit = 3

Number of ways for tens digit = 5

Number of ways for hundreds digit = 5

Number of 3 digits odd number =  3 × 5 × 5 = 75 

∴ 75 Three-digit odd numbers can be formed from the digits 5, 6, 7, 8, 9 if the digits can be repeated

In how many ways can we sort the letters of the word MANAGEMENT so that the comparative position of vowels and consonants remains the same as in MANAGEMENT.

  1. 1280
  2. 720
  3. 960
  4. 1080

Answer (Detailed Solution Below)

Option 4 : 1080

Permutation and Combination Question 13 Detailed Solution

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Given:

Word = MANAGEMENT

Calculation:

Vowel occupy 4 places then !4

∵ A and E are repeated then !4/(!2 × !2)

Consonant occupy 6 places then !6

⇒ M and N are repeated then !6/(!2 × !2)

 ∴ 

= 6 × 180 = 1080

In how many ways can the word CHRISTMAS be arranged so that the letters C and M are never adjacent?

  1. 8! × (7/2)
  2. 9! × (7/2)
  3. 8! × (9/2)
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : 8! × (7/2)

Permutation and Combination Question 14 Detailed Solution

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Given:

Different words from CHRISTMAS have to be formed.

Formula:

Words in which letters C and M are never adjacent = All cases – words having C and M together.

Calculation:

⇒ Total number of words = 9!/2! (Division of 2! As S is repeated)

Let us assume C and M to be one unit. Then, letters can be arranged in 8! Ways. C and M can be arranged in 2! Ways. Letter S is repeated, so total number of ways will be divided by 2!

⇒ Number of words with C and M adjacent= 8!/2! × 2! = 8!

Words in which letters C and M are never adjacent = 9!/2! – 8! = 8! × (9/2 - 1) = 8! × (7/2)

How many different 5 digit numbers can be formed from the digits 9, 0, 4, 1, 6?

  1. 120
  2. 90
  3. 84
  4. 96

Answer (Detailed Solution Below)

Option 4 : 96

Permutation and Combination Question 15 Detailed Solution

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Calculation:

The first digit of the number (from left) cannot be 0. So there are only 4 ways in which the first digit can be chosen.

For the second digit, we have 4 choices (excluding the number already chosen as the first digit).

Similarly, for the third digit, we have 3 choices, and for the fourth digit (from left), we have 2 choices and for the unit digit, we have only one choice.

Total 5 digit numbers = 4 × 4 × 3 × 2 × 1 = 96

∴ There can be 96 different 5 digit numbers can be formed from the digits 9, 0, 4, 1, 6

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