Question

There are three copies each of four different books. The number of ways in which they can be arranged in a shelf is

• A. $\frac{12!}{(3!{)}^4}$
• B. $\frac{12!}{(4!{)}^3}$
• C. $\frac{21!}{(3!{)}^{4}4!}$
• D. $\frac{12!}{(4!{)}^{3}3!}$

Answer 1

The correct option is A $\frac{12!}{(3!{)}^4}$

Here, we have $12$ books of $3$ books of each $4$ different book. Therefore, the required number of ways is
$=\frac{12!}{(3!{)}^4}$