Hello Aspirants. Welcome to Online Reasoning Section in AffairsCloud.com. Here we are creating question sample in coded **Permutations & Combinations**, which is common for all the competitive exams. We have included Some questions that are repeatedly asked in bank exams !!

** 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? **

A) 480

B) 360

C) 240

D) 360

E) 24

Answer & Explanation

** 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.

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

A) 120280

B) 453600

C) 360340

D) 3628800

E) None of these

Answer & Explanation

** B) 453600**

Explanation:

ALLIGATION contains 10 letters, so total 10! ways. There are 2 As, 2 Ls, 2 Is

So 10!/(2!*2!*2!)

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

A) 60

B) 30

C) 90

D) 70

E) 120

Answer & Explanation

** 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!

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

A) 720

B) 140

C) 120

D) 360

E) 210

Answer & Explanation

** B) 140**

Explanation:

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

** How many 3 digit numbers are divisible by 4? **

A) 256

B) 225

C) 198

D) 252

E) 120

Answer & Explanation

** 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_{n} = 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

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

A) 225

B) 240

C) 120

D) 160

E) 185

Answer & Explanation

**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

**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? **

A) 860

B) 1262

C) 1001

D) 1768

E) 984

Answer & Explanation

** 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}.

**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? **

A) 280

B) 420

C) 220

D) 495

E) 460

Answer & Explanation

** 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}.

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

A) 70

B) 96

C) 84

D) 48

E) 102

Answer & Explanation

** 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

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

A) 348

B) 284

C) 224

D) 336

E) None of these

Answer & Explanation

** 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

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