# 8 different items into 5 different boxes

#### Danijel

Problem Statement
How many ways are there to arrange 8 different items into 5 different boxes, each containing at least 1 item.
Relevant Equations
Number of injections, surjections, and inclusion-exclusion formulas
My teacher solved this using inclusion-exclusion formulas to count the number of surjections from a set of 8 elemets (containing items) to a set of 5 elements (containing boxes). However, I thought of a different solution. But I have a hunch it's wrong.
What I thought is to first make sure every box gets and item, so we are basically counting number of injections from a set of 5 elements(num. of boxes) to the set of 8 elements (number of items). Then we have to arrange the remaining 3 items into 5 boxes. Here we are counting the number of injections from a set of 3 elemets to the set of 5 elemets. In the end, we multiply these two results to obtain the final one.
Please, let me know what you think.

Edit: I have found a mistake in my reasoning. This only makes sure that every box gets at most 2 items. Which doesn't have to be true. One box can contain up to 4 items, respecting the problem setttings.

Last edited:
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#### SammyS

Staff Emeritus
Homework Helper
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Problem Statement: How many ways are there to arrange 8 different items into 5 different boxes, each containing at least 1 item.
Relevant Equations: Number of injections, surjections, and inclusion-exclusion formulas

My teacher solved this using inclusion-exclusion formulas to count the number of surjections from a set of 8 elements (containing items) to a set of 5 elements (containing boxes). However, I thought of a different solution. But I have a hunch it's wrong.
What I thought is to first make sure every box gets and item, so we are basically counting number of injections from a set of 5 elements(num. of boxes) to the set of 8 elements (number of items). Then we have to arrange the remaining 3 items into 5 boxes. Here we are counting the number of injections from a set of 3 elements to the set of 5 elements. In the end, we multiply these two results to obtain the final one.
Please, let me know what you think.

Edit: I have found a mistake in my reasoning. This only makes sure that every box gets at most 2 items. Which doesn't have to be true. One box can contain up to 4 items, respecting the problem setttings.
Yes, you have found your mistake. That's good thinking !

Here's a idea.
For arranging the remaining 3 items into 5 boxes, would it work to simply find the number of functions from a set of 3 elements to a set of 5 elements? (I think you would have then account for some duplicate counting.)

#### PeroK

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2018 Award
Another approach is to count the possibilities for each pattern. There are only three:

4, 1, 1, 1, 1
3, 2, 1, 1, 1
2, 2, 2, 1, 1

#### PeroK

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2018 Award
Yet another approach is to generate the iterative sequence (which is actually on OEIS).

If we let $X_n$ be the number of ways of putting 8 different objects into $n$ non-empty boxes, then:

$X_1 = 1$
$X_2 = 2^8 - 2X_1$
$X_3 = 3^8 - 3X_2 - 3X_1$

Etc.

#### LCKurtz

Homework Helper
Gold Member
@Danijel : So what did you get for your answer?

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