# Permutations And Combinations - MCQ 3

## 20 Questions MCQ Test Quantitative Aptitude for Competitive Examinations | Permutations And Combinations - MCQ 3

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This mock test of Permutations And Combinations - MCQ 3 for Quant helps you for every Quant entrance exam. This contains 20 Multiple Choice Questions for Quant Permutations And Combinations - MCQ 3 (mcq) to study with solutions a complete question bank. The solved questions answers in this Permutations And Combinations - MCQ 3 quiz give you a good mix of easy questions and tough questions. Quant students definitely take this Permutations And Combinations - MCQ 3 exercise for a better result in the exam. You can find other Permutations And Combinations - MCQ 3 extra questions, long questions & short questions for Quant on EduRev as well by searching above.
QUESTION: 1

### In how many ways 5 rings can be worn on 3 fingers?

Solution:

C) 60
Explanation: 0 0 0
Let these 3 circles are 3 fingers For 1st finger we have 5 choices, for second finger we have 4 choices left of rings, for third finger we have 3 choices left.
So total 5*4*3 = 60 ways

QUESTION: 2

### In how many ways the letters of the word ‘AUTHOR’ be arranged taking all the letters?

Solution:

B) 720
Explanation: AUTHOR contains 6 letters, so total 6! ways.

QUESTION: 3

### In how many ways the letters of the word ‘MINIMUM’ be arranged taking all the letters?

Solution:

A) 420
Explanation: MINIMUM contains 7 letters, so total 7! ways. But it contains 2 I’s and 3 M’s so divide by 2! And 3!
So ways 7!/(2! * 3!) = 7*6*5*4*3*2*1 / 2*1*3*2*1 = 420

QUESTION: 4

How many words of 4 letters with or without meaning be made from the letters of the word ‘LEADING’, when repetition of letters is allowed?

Solution:

D) 57624
We have 4 places where letters are to be placed.
For first letter there are 7 choices, since repetition is allowed, for second, third and fourth letter also we have 7 choices each, so total of 7*7*7*7 ways = 2401 ways.
Now for arrangement of these 4 words, we have 4! Ways.
So total of 2401 * 4! Ways.

QUESTION: 5

In how many ways letters of word ‘INVISIBLE’ be arranged such that all vowels are together?

Solution:

B) 2880
Explanation: First make IIIE in a circle. So we have Now we have N, V, S, B, L and box, their arrangements can be done in 6!
Letters inside circle are also to be arranged, we have I, I, I, E so ways are 4!/3!
Total ways 6! * 4!/3!

QUESTION: 6

How many words can be made out of the letters of word ‘POUNDING’ such that all vowels occupy odd places?

Solution:

A) 1440
Explanation: In POUNDING, there are 8 places 1 2 3 4 5 6 7 8
So for 3 places selection of vowels, we have 1, 3, 5, 7 number places 4C3 ways Now for arranging these 3 vowels, ways are 3!
Remaining 5 are consonants (in which there are 2 N’s) for which 5!/2! so total ways = 4C3 *3!*(5!/2!)

QUESTION: 7

In how many ways a group of 2 men and 4 women be made out of a total of 4 men and 7 women?

Solution:

B) 210
Explanation: We have to select 2 men from 4 men, and 4 women from 7 women So total ways = 4C2 * 7C4

QUESTION: 8

There are 8 men and 7 women. In how many ways a group of 5 people can be made such that at least 3 men are there in the group?

Solution:

C) 1722
Explanation: Case 1: 3 men and 2 women 8C3 * 7C2 = 1176
Case 2: 4 men and 1 women 8C4 * 7C1 = 490
Case 3: all 5 men 8C= 56

QUESTION: 9

There are 6 men and 7 women. In how many ways a committee of 4 members can be made such that a particular woman is always included.

Solution:

D) 220
Explanation: There are total 13 people, a particular woman is to be included, so now 12 people are left to chosen from and 3 members to be chosen. So ways are 12C3 .

QUESTION: 10

There are 5 men and 3 women. In how many ways a committee of 3 members can be made such that 2 particular men are always to be excluded.

Solution:

B) 20
Explanation: Total 8 people, 2 men are to excluded, so 6 men left to be chosen from and 3 members to be chosen. So ways are 6C3 .

QUESTION: 11

In a exam paper 1st section contains 10 questions each with 5 choices and  second section contains 5 questions each with 4 choices. In how manydifferent ways can the paper be answered if all the questions are attempted.

Solution:

Answer – B. 510 × 45
Explanation :
10 questions with 5 choices = 510
20 questions with 4 choices =  45

QUESTION: 12

How many four digit number can be formed with the digits 5,9,1 and 3 only ?

Solution:

Answer – C.256 Explanation : 4*4*4*4 = 256

QUESTION: 13

Find the number of ways in which 9 students equally divided into three groups ?

Solution:

Answer – A.280 Explanation : 9!/[3! ×(3!)3 ] = 362880/6*216 = 280

QUESTION: 14

Find the number of ways of distributing 9 identical balls into 4 boxes so that no box is empty and each box being large enough to accommodate all balls ?

Solution:

Answer – D.56 Explanation :  9-1C4-18C3 = 8*7*6/3*2*1 = 56

QUESTION: 15

12 students participated in  the competition and each get different score. In how many ways can three different prizes given ?

Solution:

Answer – A.1320 Explanation : 12*11*10 = 1320

QUESTION: 16

How many arrangement can be made from the word COMMERCE, such that all the vowels do not come together ?

Solution:

Answer – D.9000 Explanation : 8 letters = 8!/ 2! 2! = 40320/4 = 10080 6 letters = 6!/2! = 360 Vowels = 3!/2! =  3 No of ways vowels together = 360*3 = 1080 No of ways vowels not together = 10080 – 1080 = 9000

QUESTION: 17

Three brothers have 5 shirts, 8 pants and 6 ties. In how many ways can they wear them ?

Solution:

Answer – C. 2419200 Explanation :  5P3 * 8P3 *   6P3 = 60*336*120 = 2419200

QUESTION: 18

5 men and 3 women are to be seated such that no 2 women sit together and  2 men sit together. Find the no of ways in which this can be arranged ?

Solution:

Answer – B.720 Explanation : 5!*3! = 120*6 = 720

QUESTION: 19

A group consists of 3 couples  in which each of the 3 men have one wife each.In how many ways could they arranged in a straight line so that the men and women occupy alternate position ?

Solution:

Answer – D.72 Explanation : 3!*3! + 3!*3! = 36+36 = 72

QUESTION: 20

In how many ways can 5 different balls be distributed to 4 different boxes, when each of the can hold any no of ball ?

Solution:

Answer – A.1024 Explanation : No of way =45 = 1024