A library has `a` copies of one book, `b` copies each of two books, `c` copies each of three books, a single copy of `d` books. The total number of ways in which these books can be arranged in a shelf is equal to a. `((a+2b+3c+d)!)/(a !(b !)^2(c !)^3)` b. `((a+2b+3c+d)!)/(a !(2b !)^(c !)^3)` c. `((a+b+3c+d)!)/((c !)^3)` d. `((a+2b+3c+d)!)/(a !(2b !)^(c !)^)`

Options

A. `((a+b+c+d)!)/(a!b!c!)`
B. `((a+2b+3c+3d)!)/(a!(b!)^(2)(c!)^(3))`
C. `((2+2b+3c+d)!)/(a!b!c!)`
D. none

Answer

B