• # question_answer 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 books 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

Total number of books $=a+2b+3c+d$ Since there are $b$ copies of each of two books, $c$ copies of each of three books and single copies of $d$ books. Therefore the total number of arrangements is $\frac{(a+2b+3c+d)\ !}{a\ !\ {{(b\ !)}^{2}}{{(c\ !)}^{3}}}$.