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Applications of Permutations And Combinations - Commerce MCQ


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10 Questions MCQ Test - Applications of Permutations And Combinations

Applications of Permutations And Combinations for Commerce 2024 is part of Commerce preparation. The Applications of Permutations And Combinations questions and answers have been prepared according to the Commerce exam syllabus.The Applications of Permutations And Combinations MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Applications of Permutations And Combinations below.
Solutions of Applications of Permutations And Combinations questions in English are available as part of our course for Commerce & Applications of Permutations And Combinations solutions in Hindi for Commerce course. Download more important topics, notes, lectures and mock test series for Commerce Exam by signing up for free. Attempt Applications of Permutations And Combinations | 10 questions in 10 minutes | Mock test for Commerce preparation | Free important questions MCQ to study for Commerce Exam | Download free PDF with solutions
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 for Applications of Permutations And Combinations - 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 for Applications of Permutations And Combinations - 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 for Applications of Permutations And Combinations - 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 for Applications of Permutations And Combinations - 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 for Applications of Permutations And Combinations - 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 for Applications of Permutations And Combinations - 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 for Applications of Permutations And Combinations - 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 for Applications of Permutations And Combinations - 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 for Applications of Permutations And Combinations - 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 for Applications of Permutations And Combinations - 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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