Question

We have 4different boxesand 6different objects. We want to distribute the objects into the boxes such that at no box is empty. In how many ways can this be done?

Answer 1

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 $(r-k)$ objects. All can be put in any of k boxes.  Therefore, these $(r-k)$ objects could in the $k^{(r-k)}$ manner are organized.  Consequently, both possible ways to do this are

$=\binom{r}{k} \times k! \times k^{r-k}\\\\=\frac{r! \times k^{r-k}}{(r-k)!}$

Consequently, the number of ways that r objects in k different boxes can be arranged to make no book empty is every possible one

$= \frac{r!k^{r-k}}{(r-k)!}$