Q. A library has a copies of one book, $b$ copies of each of two books, $c$ copies of each of three books and single copies of $d$ books. The total number of ways in which these book can be distributed, 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
Explanation
(b)
Total number of books $= a + 2b + 3c + d$
$\therefore$ The Total number of arrangements
$= \frac{(a + 2b + 3c + d)!}{a!(b!)^{2}(c!)^{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.