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
Correct Answer: B
Solution :
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 \[\frac{(a+2b+3c+d)\ !}{a\ !\ {{(b\ !)}^{2}}{{(c\ !)}^{3}}}\].
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