There are 20 books of which 4 are single volumes and the other are books of 8, 5 and 3 volumes respectively. In how many ways can all these books be arranged on a self so that volumes of the same book are not separated?

Possible arrangements are 7!/2! 2! = 1260
[Division by 2 times 2! is because of the repletion of E and R .]

Taping all vowels (IEO) as a single letter (since they come together ) there are six letters with two 'R' s
Hence no. of arrangement = 6!/2! x 3! = 2160
[3 vowels can be arranged in 3! ways among themselves, here multiplied with 3!. ]

There are 12 letters in the world 'civilization' of which four are i's and other are different letters.
? Total number of permutations = 12!/4!
But one word is civilization itself.
? Required number of rearrangements = 12!/4! - 1

Let there be n persons in the room. The total number of hand shakes is same as the number of ways of selecting 2 out of n.
ⁿC₂ = 66
? n(n - 1) / 2! = 66
? n² - n - 132 = 0
? (n - 12) (n + 11) = 0
? n = 12

There are 3A's 2N's and one B. We have to find the total number of arrangements of 6 letters out of which 3 are alike of one kind, 2 are alike of second kind, thus the total number of words
= 6! / (3! 2!) = 60

Regarding all copies of the same book as one book, we have only 5 books. These 5 books can be arranged in 5! ways. But all copies of the same book being identical can be arranged in only one way.

? Required number = 5! x 1! x 1! x 1! x 1! = 120

Assume the 2 girl students to be together i,e (one). Now there are 5 students.
Possible ways of arranging them are 5! = 120
Now they (two girl) can arrange themselves in 2! ways.

Hence, total ways = 120 x 2! = 240

The required number of words is : (²C₁ x ⁴C₂ + ²C₂ x ²C₁) 3! = 96

Number of ways of selecting one or more friends from 5 friends
= ⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅
= 5 + 10 + 10 + 5 + 1
= 31 ways
Number of ways of selecting one or more friends from 4 friends = ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄
= 4 + 6 + 4 + 1
= 15 ways
? Total number of ways = 31 + 15 = 46 ways

Three numbers can be selected and arranged out of 10 numbers in ¹⁰P₃ ways 10!/7! = 10 x 9 x 8
Now, this arrangement is restricted to a given condition that first number is always less than the second number and second number is always than the third number. Thus, three numbers can be arranged among themselves in 3! ways.
Hence, required number of arrangement = (10 x 9 x 8)/(3 x 2)
= 120 ways