# Aptitude Questions: Permutations & Combinations Set 4

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 !!

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?
A) 480
B) 360
C) 240
D) 360
E) 24
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.

2. 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
B) 453600
Explanation:

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

3. 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
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!

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?
A) 720
B) 140
C) 120
D) 360
E) 210
B) 140
Explanation:

Total ways = 8C4*5C3

5. How many 3 digit numbers are divisible by 4?
A) 256
B) 225
C) 198
D) 252
E) 120
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, an = 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

6. How many 3 digits numbers have exactly one digit 2 in the number?
A) 225
B) 240
C) 120
D) 160
E) 185
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

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?
A) 860
B) 1262
C) 1001
D) 1768
E) 984
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 14C4.

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?
A) 280
B) 420
C) 220
D) 495
E) 460
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 12C4.

9. 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
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

10. 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
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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