Question

A library has 'a' copies of one book, 'b' copies each of two books, 'c' copies each of three books, and single copy of 'd' books. The total number of ways in which these books can be arranqed in a shelf is equal to

• A. $\frac{(a+2b+3c+d)!}{a!(b!{)}^2(c!{)}^3}$
• B. $\frac{(a+2b+3c+d)!}{a!(2b!{)}^2(c!{)}^3}$
• C. $\frac{(a+b+3c+d)!}{(c!{)}^3}$
• D. $\frac{(a+2b+3c+d)!}{a!\left(2b\right)!\left(3c\right)!}$

Answer 1

The correct option is A $\frac{(a+2b+3c+d)!}{a!(b!{)}^2(c!{)}^3}$

This is a general question of, arrangement of books or some material in different groups.

Total books $a+2b+3c+d$

Number of ways $(a+2b+3c+d)!$

However removing repeatetion we have $\frac{(a+2b+3c+d)!}{a!\left(b!{)}^2\right(c!{)}^3}$

Hence, the answer is $\frac{(a+2b+3c+d)!}{a!\left(b!{)}^2\right(c!{)}^3}.$