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QUESTION: 1

How many words of 4 letters with or without meaning be made from the letters of the word ‘NUMBER’, when repetition of letters is not allowed?

Solution:

D) 360

Explanation: NUMBER is 6 letters.

We have 4 places where letters are to be placed.

For first letter there are 6 choices, since repetition is not allowed, for second, third and fourth letter also we have 5, 4, and 3 choices resp., so total of 6*5*4*3 ways = 360 ways.

QUESTION: 2

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

Solution:

B) 453600

Explanation: ALLIGATION contains 10 letters, so total 10! ways. There are 2 As, 2 Ls, 2 Is So 10!/(2!*2!*2!)

QUESTION: 3

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

Solution:

A) 60

Explanation: Take vowels in a box together as one – IIU, M, N, M, M So there are 5 that to be placed for this 5!, now 3 Ms, so 5!/3!, so arrangement of vowels inside box gives 3!/2!

So total = 5!/3! * 3!/2!

QUESTION: 4

In how many ways a group of 4 men and 3 women be made out of a total of 8 men and 5 women?

Solution:

B) 140

Explanation: Total ways = ^{8}C_{4} * ^{5}C_{3}

QUESTION: 5

How many 3 digit numbers are divisible by 4?

Solution:

B) 225

Explanation: A number is divisible by 4 when its last two digits are divisible by 4 For this the numbers should have their last two digits as 00, 04, 08, 12, 16, … 96 By the formula, a = a + (n-1)d 96 = 0 + (n-1)*4 n = 25 so there are 25 choices for last 2 digits and 9 choices (1-9) for the 1st digit so total 9*25

QUESTION: 6

How many 3 digits numbers have exactly one digit 2 in the number?

Solution:

A) 225

Explanation: 0 cannot be placed at first digit to make it a 3 digit number. 3 cases: Case 1: 2 is placed at first place 1 choice for the first place, 9 choices each for the 2nd and 3rd digit (0-9 except 2) So numbers = 1*9*9 = 81 Case 2: 2 is placed at second place 8 choices for the first place (1-9 except 2), 1 choice for the 2nd digit and 9 choices for the 3rd digit (0-9 except 2) So numbers = 8*1*9 = 72 Case 3: 2 is placed at third place 8 choices for the first place (1-9 except 2), 9 choices for the 2nd digit (0-9 except 2) and 1 choice for the 3rd digit So numbers = 8*9*1 = 72 So total numbers = 81+72+72 = 225

QUESTION: 7

There are 8 men and 7 women. In how many ways a group of 5 people can be made such that the particular woman is always to be included?

Solution:

C) 1001

Explanation: Total 15 people, and a particular woman is to be taken to form a group of 5, so choice is to be done from 14 people of 4 people Ways are ^{14}C_{4} .

QUESTION: 8

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

Solution:

D) 495

Explanation: There are total 13 people, a particular man is to be excluded, so now 12 people are left to chosen from and 4 members to be chosen. So ways are ^{12}C_{4} .

QUESTION: 9

How many 4 digit words can be made from the digits 7, 8, 5, 0, and 4 without repetition?

Solution:

B) 96

Explanation: 0 cannot be on first place for it to be a 4 digit number, So for 1st digit 4choices, for second also 4 (because 0 can be placed here), then 3 for third place, 2 for fourth place Total numbers = 4*4*3*2

QUESTION: 10

In how many ways 8 students can be given 3 prizes such that no student receives more than 1 prize?

Solution:

D) 336

Explanation: For 1st prize there are 8 choices, for 2nd prize, 7 choices, and for 3rd prize – 6 choices left So total ways = 8*7*6

QUESTION: 11

In how many ways can 3 prizes be given away to 12 students when each student is eligible for all the prizes ?

Solution:

Answer – B.1728 Explanation : 12^3 = 1728

QUESTION: 12

Total no of ways in which 30 sweets can be distributed among 6 persons ?

Solution:

Answer – A. ^{35}C_{5}

^{30+6-1}C_{6-1} = ^{35}C_{5}

QUESTION: 13

A bag contains 4 red balls and 5 black balls. In how many ways can i make a selection so as to take atleast 1 red ball and 1 black ball ?

Solution:

Answer – C.465 Explanation : 2^{4} -1 = 16 -1 = 15

2^{5} - 1 = 32 -1 = 31

15*31 = 465

QUESTION: 14

In how many ways can 7 beads be strung into necklace ?

Solution:

Answer – D.360 Explanation : No of way in Necklace = (n-1)!/2 = 6!/2 = 720/2 = 360

QUESTION: 15

Find the no of 3 digit numbers such that atleast one of the digit is 6 (with repetitions) ?

Solution:

Answer – A.252 Explanation : Total no of 3 digit number = 9*10*10 = 900 No of 3 digit number- none of the digit is 6 = 8*9*9 = 648 No of 3 digit number – atleast one digit is 6 = 900-648 = 252

QUESTION: 16

In how many ways can 7 girls and 4 boys stand in a row so that no 2 boys are together ?

Solution:

Answer – A. 8467200

Explanation : No of ways = 7!* ^{8}P_{4}

7! = 5040

^{8}P_{4} = 8*7*6*5 = 1680

No of ways = 5040*1680 = 8467200

QUESTION: 17

In how many ways the letters of the word PERMUTATION be arranged ?

Solution:

Answer – D. 11!/2!

Explanation : No of ways = 11!/2!

QUESTION: 18

How many numbers can be formed with the digits 1, 7, 2, 5 without repetition ?

Solution:

Answer – C.64 Explanation : 1 digit number = 4 2 digit no = 4*3 = 12 3 digit no = 4*3*2 = 24 4 digit no = 4*3*2*1 = 24 Total = 4+12+24+24 = 64

QUESTION: 19

There are 3 boxes and 6 balls. In how many ways these balls can be distributed if all the balls and all the boxes are different?

Solution:

Answer – C.729 Explanation : 3^6 = 729

QUESTION: 20

In how many ways can 4 books be selected out of 10 books on different subjects ?

Solution:

Answer – A.210 Explanation : ^{16}C_{4} = 10*9*8*7/4*3*2*1 = 5040/24 = 210

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