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So, lets say I have $4$ burglars, and I have a custom pistol with $10$ bullets in it. How many possible ways can I put $10$ bullets into EXACTLY $2$ burglars and let the other $2$ get away?

Note that I am NOT looking for an ANSWER, I am only asking if it is possible for someone to help me understand why my initial conditions are incorrect. In particular, can someone confirm if what I am doing below is correct?

*My technique is to use the inclusion exclusion formula where it is

$$E_2 = S_2 - \binom{3}{1}S_3 + \binom{4}{2}S_4$$

where $S_i$ stands for the possibilities where $i$ burglars get a bullet in them.

Now, lets say I want to find $S_2$, which is the number of ways to distribute $10$ bullets amongst $2$ burglars. Would the answer to this perhaps be

$$\binom{4}{2}(10^2 - 2)$$

where we first choose $2$ burglars out of the $4$ and then with the $2$ burglars each can have $10$ bullets so $10^2$ and then I minus $2$ because each has to have at least $1$.

Can someone please confirm if I did $S_2$ correctly?

The answer is $6132$ just in case anyone is wondering...

N. F. Taussig
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Belphegor
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  • Please see this tutorial on how to typeset mathematics on this site. Also, if you right click on an expression, you can see how I formatted it by checking the LaTeX commands. – N. F. Taussig Apr 07 '15 at 09:10

1 Answers1

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As far as I can see, your explanation for $S_2$ is correct, but in your answer you wrote $10^2$ when it should be $2^{10}$.

Also, it seems to me that this should be the entire answer, you don't need to consider $S_3$ and $S_4$.

David
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