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# Applications Of Permutations And Combinations

## 10 Questions MCQ Test Mathematics (Maths) Class 11 | Applications Of Permutations And Combinations

Description
This mock test of Applications Of Permutations And Combinations for JEE helps you for every JEE entrance exam. This contains 10 Multiple Choice Questions for JEE Applications Of Permutations And Combinations (mcq) to study with solutions a complete question bank. The solved questions answers in this Applications Of Permutations And Combinations quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Applications Of Permutations And Combinations exercise for a better result in the exam. You can find other Applications Of Permutations And Combinations extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### The number of ways in which three different rings can be worn in four fingers with at most one in each finger, are

Solution:

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

QUESTION: 2

### 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?

Solution:

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.

QUESTION: 3

### The total number of ways of answering 5 objective questions, each question having four choices are

Solution:

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

QUESTION: 4

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?

Solution:

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

QUESTION: 5

The measure of an interior angle of a regular polygon is 140°. The number of sides and diagonals in this polygon are:

Solution:

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.

QUESTION: 6

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?

Solution:

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

QUESTION: 7

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?

Solution:

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

QUESTION: 8

Four alphabets A, M, P, O are purchased from a warehouse. How many ordered pairs of initials can be formed using these?

Solution:

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

QUESTION: 9

How many words beginning with ‘T’ and ending with ‘E’ can be formed using the letters of the word”TRIANGLE” ?

Solution:

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

QUESTION: 10

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?

Solution:

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