Total number of books = a + 2b + 3c + d . Since there are 'b' copies of each of two books, 'c' copies of each of three books and single copy of 'd' book.
Therefore, the total number of arrangements is = (a + 2b + 3c + d )! / {a! (b!)² (c!)³}

Correct Option: B

Total number of books = a + 2b + 3c + d . Since there are 'b' copies of each of two books, 'c' copies of each of three books and single copy of 'd' book.
Therefore, the total number of arrangements is = (a + 2b + 3c + d )! / {a! (b!)² (c!)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!

Correct Option: B

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

Taping all vowels (IEO) as a single letter (since they come together ) there are six letters with two 'R' s

Correct Option: C

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!. ]

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

Correct Option: C

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

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.

Correct Option: A

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!