A library has two books each having three copies and three other books each having two copies. In how many ways can all these books be arranged in a shelf so that copies of the same book are not separated?

Volumes of the same book are not to be separated i, e. all volumes of the same book are to be kept together, Regarding all volumes of the same book as one book, we have only 4 + 1 + 1 + 1 = 7 books.

These seven books can be arranged in 7! ways. The book having 8 volumes can be arranged among themselves in 8! ways, the book having 5 volumes can be arranged among themselves in 5! ways, and the book having 3 volumes can be arranged among themselves in 3! ways.

? Required number = 7! 8! 5! 3!

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

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

Total number of password using all alphabets -Total
number of password using no symmetric alphabets
= (26 x 25 x 24 ) - (15 x 14 x 13 )
= 12870