Question

A library has $a$ copies of one book, $b$ copies of two books, $c$ copies of each of three books and a single copy of $d$books. The total number of ways in which these books can be distributed is?

• A. $\frac{(a+2b+3c+d)!}{a!\times b!\times c!}$
• B. $\frac{(a+2b+3c+d)!}{a!\times b^2!\times c^3!}$
• C. $\frac{(a+2b+3c+d)!}{a!\times {(b!)}^2\times {(c!)}^3}$
• D. $\frac{(a+2b+3c+d)!}{a+b+c+d}$

Answer 1

The correct option is C

$\frac{(a+2b+3c+d)!}{a!\times {(b!)}^2\times {(c!)}^3}$

Explanation for the correct option:

Finding the total number of ways books can be distributed:

Total number of books$=a+2b+3c+d$

Therefore the number of ways they can arranged $=(a+2b+3c+d)!$

Number repeated books are $b\&c$ $2\&3$ times respectively.

Therefore total number of arrangement without repetition

$\frac{(a+2b+3c+d)!}{a!\times {(b!)}^2\times {(c!)}^3}$

Hence, correct option is $\mathbf{\left(}\mathit{C}\mathbf{\right)}\mathbf{.}$