Question

13. A book publisher has 3000 copies of a discrete mathematics book. How many ways are there to store these books in their three warehouses if the copies of the book are indistinguishable?

Short answer

There are $4.504.501$different ways to store these books in their three warehouses if the copies of the book are indistinguishable

Step-by-step solution

Definitions

Definition of Permutation (Order is important)

No repetition allowed:$P(n,r)=\frac{n!}{(n-r)!}$

Repetition allowed:$n^T$

Definition of combination (order is important)

No repetition allowed:$C(n,r)=\left(\begin{array}{l}n\\ r\end{array}\right)=\frac{n!}{r!(n-r)!}$

Repetition allowed:$C(n+r-1,r)=\left(\begin{array}{c}n+r-1\\ r\end{array}\right)=\frac{(n+r-1)!}{r!(n-1)!}$

with$n!=n(n-1)\cdot .....\cdot 2\cdot 1$

Step 2: Solution

The order of the elements does not matters (since the books are indistinguishable), thus we need to use the definition of combination.

We are interested in selecting r = 3000 elements from a set with n = 3 elements.

Repetition of elements is allowed

$\begin{array}{r}C(n+r-1,r)=C(3+3000-1,3000)\\ \\ C(3002,3000)=\frac{3002!}{3000!(3002-3000)!}=\frac{3002!}{3000!2!}=4,504,501\end{array}$

There are $4,504,501$different ways to store these books in their three warehouses if the copies of the book are indistinguishable.