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I have encountered two problems that are almost identical. Consider this: $12$ basketballs numbered 1 through 12 are thrown at random into 20 boxes. I want to know the total number of possible outcomes (assume order matters). The answer is $20^{12}$. In the other problem (which I do not need to mention), follows the same pattern and the answer is also $n^k$, where $n$ is the number of slots and $k$ is the number of objects I have. I came up with the $n^k$ conclusion by myself.

I am just wondering, why is this the number of possible outcomes?

Here are my efforts to understand it:

  1. I Googled the problem, but I got no results.

  2. I wanted to confirm that this $n^k$ conclusion is true. I did a simpler example of 2 objects and 4 slots and the result was matching the $n^k$ form.

  3. I tried to use the multiplication principle (the counting one). We can say this: first stage throw all balls each basketball gets into a different box. It is Perms(20, 12). Well, there is no a second stage. If I want to "add" more ways, I need to use the counting principle again in another context. Let us say make two balls in each box... etc. It is just a crazy overhead counting.

  4. I tried to make an analogy with the problem of rolling two six-sided fair dice where the total outcomes of the pair they make is $6^6$. But it did not click with this basketball problem.

Notes:

  1. You can throw more than one basketball (at random) into one box.

  2. This problem was encountered in a "probability stuff" context. I was able to go through the numerator but not this one.

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2 Answers 2

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First, you throw the first ball (the ball numbered 1). There are $n$ possible outcomes for this

Second, you throw the numbered 2 ball, there are also $n$ possible outcomes

and so on ...

Finally, you throw the last ball (the k-th ball), there are also $n$ possible outcomes

You use multiplication principle then you get the anwser is $n \times n \times ... \times n = n^k$ possible outcomes

Is that anwser your question?

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  • $\begingroup$ I have not thought of it this way at all .. I mean the idea that each stage is basically throwing ONE ball at a time. $\endgroup$
    – Gold_Sky
    Commented Jun 10, 2016 at 2:47
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What did you google?

The key word you need is combinatorix. Specifically, combinatorix counting. This site has a decent description of why.

https://en.wikibooks.org/wiki/Probability/Combinatorics

I hope it helps.

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  • $\begingroup$ I googled the problem itself. I have not thought of googling n^k. $\endgroup$
    – Gold_Sky
    Commented Jun 10, 2016 at 2:50

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