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Applications of Permutations And Combinations - Free MCQ Test with solutions


MCQ Practice Test & Solutions: Applications of Permutations And Combinations (10 Questions)

You can prepare effectively for JEE Mathematics (Maths) for JEE Main & Advanced with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Applications of Permutations And Combinations ". These 10 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 10

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Applications of Permutations And Combinations - Question 1
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The number of ways in which three different rings can be worn in four fingers with at most one in each finger, are

Detailed Solution: Question 1

The total number of ways is same as the number of arrangements of 4 fingers, taken 3 at a time.
So, required number of ways = 4P3 
= 4!/(4-3)!
= 4!/1!
= 4! => 24

Applications of Permutations And Combinations - Question 2
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A room has 8 doors. In how many ways, a man can enter in the room through one door and exit through a different door?

Detailed Solution: Question 2

The person has 8 options to enter the hall. For each of these 8 options, he has 7 options to exit the hall. Thus, he has 8 × 7 = 56 ways to enter and exit from different doors.

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Applications of Permutations And Combinations - Question 3
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The total number of ways of answering 5 objective questions, each question having four choices are

Detailed Solution: Question 3

There are five questions
Each question has 4 options
No. of possible ways of answering each question is four
No. of Possible ways for Q1 = 4
No. of Possible ways for Q2 = 4
No. of Possible ways for Q3 = 4
No. of Possible ways for Q4 = 4
No. of Possible ways for Q5 = 4
So, Total number of ways of answering 5 objective type questions, each question having 4 choices = 45
= 1024

Applications of Permutations And Combinations - Question 4
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In how many ways, a party of 5 men and 5 women be seated at a circular table, so that no two women are adjacent?

Detailed Solution: Question 4

Lets first place the men (M). '*' here indicates the linker of round table
 
* M -M - M - M - M *
which is in (5-1)! ways
So we have to place the women in between the men which is on the 5 empty seats ( 4 -'s and 1 linker i.e * )
So 5 women can sit on 5 seats in (5)! ways or
1st seat in 5 ways
2nd seat 4
3rd seat 3
4th seat 2
5th seat 1
i.e 5*4*3*2*1 ways
So the answer is 5! * 4! = 2880

Applications of Permutations And Combinations - Question 5
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The measure of an interior angle of a regular polygon is 140°. The number of sides and diagonals in this polygon are:

Detailed Solution: Question 5

Since the exterior angle 140 degrees, The sum of the interior angles = (2n - 4)* right angles. So 140n = (2n - 4)* right angles, or
140n = (2n - 4)*90, or
140n = 180n - 360o, or
40n = 360°, or
n = 9 sides.

Applications of Permutations And Combinations - Question 6
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In how many ways can a cricket team of 11 players be chosen out from a squad of 14 players, if 5 particular players are always chosen?

Detailed Solution: Question 6

Total no of players = 14 out of which 5 are fixed.
So, 11-5 = 6
Remaining players = 14 - 6
= 9 players
9C6 = 9!/(3!*6!)
= 84 

Applications of Permutations And Combinations - Question 7
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A team of 7 players is to be formed out of 5 under 19 players and 6 senior players. In how many ways, the team can be chosen when at least 4 senior players are included?

Detailed Solution: Question 7

No. of ways to select 4 senior and 3 U-19 players = 6C4 * 5C3 = 150
No. of ways to select 5 senior and 2 U-19 players = 6C5 * 5C2 = 60
No. of ways to select 6 senior and 1 U-19 players = 6C6 * 5C1 = 5 
Total no. of ways to select the team = 150 + 60 + 5 = 215

Applications of Permutations And Combinations - Question 8
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Four alphabets A, M, P, O are purchased from a warehouse. How many ordered pairs of initials can be formed using these?

Detailed Solution: Question 8

Total number of letters = 4
 Number of ordered pairs of letters that can be formed like (A, M) or (P, O) etc = 4P2 ​= 4!/2!
​= 24/2
​= 12

Applications of Permutations And Combinations - Question 9
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How many words beginning with ‘T’ and ending with ‘E’ can be formed using the letters of the word”TRIANGLE” ?

Detailed Solution: Question 9

There are 8 letters in the word TRIANGLE.
2 alphabets are fixed, remaining are 6 alphabets
So, number of arrangements = 6P6
= 6!
= 720

Applications of Permutations And Combinations - Question 10
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In how many ways can 4 red, 3 yellow and 2 green chairs be arranged in a row if the chairs of the same colour are indistinguishable?

Detailed Solution: Question 10

Total no of balls = 9
red balls = 4
yellow balls = 3
green balls = 2
Total no. of arrangements = 9!/(4!*3!*2!)
= 1260

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About this JEE Practice Test

This test contains 10 MCQs on Applications of Permutations And Combinations with detailed solutions for each question. It is part of the Mathematics (Maths) for JEE Main & Advanced course on EduRev, designed to align with the current JEE syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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