Hiremee Permutation and Combinations Questions with Answers

Total number of letters = 6 and R is repeated twice.

Required arrangements = (6! / 2!) = 360

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

=> (4c3) * (4c2) = 24

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

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.

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.

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

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

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.

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.