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June 23, 2019

Question 1

720

300

640

360

None of these

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Total number of letters = 6 and R is repeated twice. Required arrangements = (6! / 2!) = 360

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Question 2

15

16

20

24

None

Total number of ways = (select 3 boys from group of 4 boys) * (2 girls from group of 4 girls) => (4c3) * (4c2) = 24

Question 3

210

1050

25200

21400

Number of ways of selecting 3 consonants from 7 = 7C3 Number of ways of selecting 2 vowels from 4 = 4C2 Number of ways of selecting 3 consonants from 7 and 2 vowels from 4 = 7C3 × 4C2 It means we can have 210 groups where each group contains total 5 letters (3 consonants and 2 vowels). Number of ways of arranging 5 letters among themselves 5!=120 Hence, required number of ways 210*120=25200

Question 4

5

10

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it. The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place. The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it. Required number of numbers = (1 x 5 x 4) = 20.

Question 5

32

48

64

96

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black). Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3) = 64.

Question 6

195052

195053

195054

185054

nPr =n!/(n−r)! 59P3=59! / (56)! =59∗58∗57∗56! / (56)! =195054

Question 7

52

72

120

We need to find the ways that vowels NEVER come together. Vowels are A, E Let the word be FTR(AE) having 4 words. Total ways = 4! = 24 Vowels can have total ways 2! = 2 Number of ways having vowel together = 48 Total number of words using all letter = 5! = 120 Number of words having vowels never together = 120-48 = 72

Question 8

12

168

This question seems to be a bit typical, isn't, but it is simplest. 1 red ball can be selected in 4C1 ways 1 white ball can be selected in 3C1 ways 1 blue ball can be selected in 2C1 ways Total number of ways = 4C1 x 3C1 x 2C1 = 4 x 3 x 2 = 24 Please note that we have multiplied the combination results, we use to add when their is OR condition, and we use to multiply when there is AND condition, In this question it is AND as 1 red AND 1 White AND 1 Blue, so we multiplied.

Question 9

10080

4989600

120960

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI). Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different. Number of ways of arranging these letters = 8! /(2!)(2!) = 10080. Now, AEAI has 4 letters in which A occurs 2 times and the rest are different. Number of ways of arranging these letters = 4! /2!= 12. Required number of words = (10080 x 12) = 120960.

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