Permutation and Combination MCQ Quiz - Objective Question with Answer for Permutation and Combination - Download Free PDF
Last updated on Jul 22, 2025
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?
Answer (Detailed Solution Below)
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 ⇒
Case 2: 3 men, 2 women ⇒
Case 3: 4 men, 1 woman ⇒
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?
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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)?
Answer (Detailed Solution Below)
Permutation and Combination Question 6 Detailed Solution
Download Solution PDF⇒ 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?
Answer (Detailed Solution Below)
Permutation and Combination Question 7 Detailed Solution
Download Solution PDFExplanation:
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
C 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?
Answer (Detailed Solution Below)
Permutation and Combination Question 8 Detailed Solution
Download Solution PDFGiven:
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 ?
Answer (Detailed Solution Below)
Permutation and Combination Question 9 Detailed Solution
Download Solution PDFGiven:
(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.
Answer (Detailed Solution Below)
Permutation and Combination Question 10 Detailed Solution
Download Solution PDF448 = 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?
Answer (Detailed Solution Below)
Permutation and Combination Question 11 Detailed Solution
Download Solution PDFGiven
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
Answer (Detailed Solution Below)
Permutation and Combination Question 12 Detailed Solution
Download Solution PDFGiven:
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.
Answer (Detailed Solution Below)
Permutation and Combination Question 13 Detailed Solution
Download Solution PDFGiven:
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?
Answer (Detailed Solution Below)
Permutation and Combination Question 14 Detailed Solution
Download Solution PDFGiven:
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?
Answer (Detailed Solution Below)
Permutation and Combination Question 15 Detailed Solution
Download Solution PDFCalculation:
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