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Test: Permutations & Combinations: Circular Arrangement(June 27) - JEE MCQ


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10 Questions MCQ Test - Test: Permutations & Combinations: Circular Arrangement(June 27)

Test: Permutations & Combinations: Circular Arrangement(June 27) for JEE 2024 is part of JEE preparation. The Test: Permutations & Combinations: Circular Arrangement(June 27) questions and answers have been prepared according to the JEE exam syllabus.The Test: Permutations & Combinations: Circular Arrangement(June 27) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Permutations & Combinations: Circular Arrangement(June 27) below.
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Test: Permutations & Combinations: Circular Arrangement(June 27) - 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

Detailed Solution for Test: Permutations & Combinations: Circular Arrangement(June 27) - 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

Test: Permutations & Combinations: Circular Arrangement(June 27) - 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?

Detailed Solution for Test: Permutations & Combinations: Circular Arrangement(June 27) - 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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Test: Permutations & Combinations: Circular Arrangement(June 27) - Question 3

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

Detailed Solution for Test: Permutations & Combinations: Circular Arrangement(June 27) - 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

Test: Permutations & Combinations: Circular Arrangement(June 27) - 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?

Detailed Solution for Test: Permutations & Combinations: Circular Arrangement(June 27) - 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

Test: Permutations & Combinations: Circular Arrangement(June 27) - Question 5

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

Detailed Solution for Test: Permutations & Combinations: Circular Arrangement(June 27) - 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.

Test: Permutations & Combinations: Circular Arrangement(June 27) - 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?

Detailed Solution for Test: Permutations & Combinations: Circular Arrangement(June 27) - 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 

Test: Permutations & Combinations: Circular Arrangement(June 27) - 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?

Detailed Solution for Test: Permutations & Combinations: Circular Arrangement(June 27) - 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

Test: Permutations & Combinations: Circular Arrangement(June 27) - Question 8

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

Detailed Solution for Test: Permutations & Combinations: Circular Arrangement(June 27) - 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

Test: Permutations & Combinations: Circular Arrangement(June 27) - Question 9

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

Detailed Solution for Test: Permutations & Combinations: Circular Arrangement(June 27) - 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

Test: Permutations & Combinations: Circular Arrangement(June 27) - 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?

Detailed Solution for Test: Permutations & Combinations: Circular Arrangement(June 27) - 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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