Answer:
Following are the solution to this question:
Step-by-step explanation:
They provide various boxes or various objects. It also wants objects to be distributed into containers, so no container is empty. All we select k objects of r to keep no boxes empty, which (r C k) could be done. All such k artifacts can be placed in k containers, each of them in k! Forms. There will be remaining [tex](r-k)[/tex] objects. All can be put in any of k boxes. Therefore, these [tex](r-k)[/tex] objects could in the [tex]k^{(r-k)}[/tex] manner are organized. Consequently, both possible ways to do this are
[tex]=\binom{r}{k} \times k! \times k^{r-k}\\\\=\frac{r! \times k^{r-k}}{(r-k)!}[/tex]
Consequently, the number of ways that r objects in k different boxes can be arranged to make no book empty is every possible one
[tex]= \frac{r!k^{r-k}}{(r-k)!}[/tex]
