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

Answer 1

The correct option is C

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

Find the total number of ways of books distributions:

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

Number of ways in which total books can be arranged $=\left(a+2b+3c+\hspace{0.17em}d\right)!$

Now, if we exclude repetitions,

$\therefore$Total number of ways of distributions $=\frac{\mathbf{(}\mathbf{a}\mathbf{}\mathbf{+}\mathbf{}\mathbf{}\mathbf{2}\mathbf{b}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}\mathbf{c}\mathbf{}\mathbf{+}\mathbf{\hspace{0.17em}}\mathbf{d}\mathbf{)}\mathbf{!}}{\mathbf{a}\mathbf{!}{\mathbf{(}\mathbf{b}\mathbf{!}\mathbf{)}}^{\mathbf{2}}{\mathbf{(}\mathbf{c}\mathbf{!}\mathbf{)}}^{\mathbf{3}}}$

Hence, Option ‘C’ is Correct.