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Luden [163]
3 years ago
14

We have 4different boxesand 6different objects. We want to distribute the objects into the boxes such that at no box is empty. I

n how many ways can this be done?
Mathematics
1 answer:
Musya8 [376]3 years ago
5 0

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)!}

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