Question

A library has a copy of one book, b copies of each of two books, c copies of each of three books and a single copy of d books. The total number of ways by which these books can be arranged is:

• A. $\frac{(a+b+c+d)!}{a!b!c!}$
• B. $\frac{(a+2b+3c+d)!}{a!(b!{)}^2(c!{)}^3}$
• C. $\frac{(a+2b+3c+d)!}{a!b!c!}$
• D. None of these

Answer 1

Answer: (A)

$\frac{(a+b+c+d)!}{a!b!c!}$

A library has 'a' copy of one book, 'b' copies of each of two books, 'c' copies of each of three books and a single copy of 'd' books

Therefore total no. of books will be $a+2b+3c+d$

we have 'a' copy of one book, 'b' copies of each of two books, 'c' copies of each of three books so these are similiar books or repeatative books.

Hence required answer will be

$\frac{(a+2b+3c+d)!}{a!(b!{)}^2(c!{)}^3}$

Answer