Question

A has library has $6$ copies of one book, $4$ copies of each of three books and single copies of $8$ books. Then number of arrangements of all the

• A. $\frac{\left(26\right)!}{(4!{)}^{3}6!}$
• B. $\frac{\left(26\right)!}{6!(4!{)}^4(6!{)}^3}$
• C. $\frac{\left(26\right)!}{6!(4!{)}^2(6!{)}^3}$
• D. $\frac{\left(26\right)!}{6!.4!.6!}$

Answer 1

The correct option is A $\frac{\left(26\right)!}{(4!{)}^{3}6!}$

The library has $6$ copies of one book, $4$ copies of each of three books and single copies of $8$ books.

$\therefore$ Total no. of books $=6\times 1+4\times 3+1\times 8=6+12+8=26$

Since the library has 6 copies of one book, $4$ copies of each of three books and single copies of 8 books.

$\therefore$ Number of ways of arranging these books $=\frac{\left(26\right)!}{(4!)(4!)(4!)(6!)}=\frac{\left(26\right)!}{{(4!)}^{3}6!}$

Hence the correct answer is $\frac{\left(26\right)!}{{(4!)}^{3}6!}$.