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I want to distribute n labeled balls into m labeled boxes. I know one obtains the number by $m^n$. But I don't quite understand why. The underlying argument is always I have m choices for the first ball m choices for the second and so on. As an example lets take 3 balls labeled A,B,C and two boxes 1,2

Now if I make all combinations I can find much more than $2^{3}=8$ possibilities. As I understand the $m^{n}$ counting procedure neglects the permutations of the n balls and the permutations of the m boxes. For example taking the configuration $B_{1}(A,B,C)$ and $B_{2}()$, then why do I not count for example $B_{1}(C,B,A)$ and $B_{2}()$ as a distinct configurations. I was sketching all possibilities and this results in $m^{n}*m!*n!$. I mean this is clearly wrong based on google searches but why? I am really stuck with this problem

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  • $\begingroup$ If you additionally want to distinguish say the order in which the boxes stay on a shelf, or the order in which the balls lie inside a box, then it is simply a different problem than the original one. $\endgroup$ Oct 1, 2018 at 9:02
  • $\begingroup$ @Michal Adamaszek. Okay so the formulation of this problem would then be. How to distribute n labeled balls into m labeled boxes by taking into account the permutations of particles and boxes?. But why does one assume to not take into account the permutations to the number of possibilities if the problem is formulated like I did in the question. I am sorry but this is confusing me so much $\endgroup$
    – zodiac
    Oct 1, 2018 at 9:08
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    $\begingroup$ Because a box with number 1 is always a box with number 1, regardless of whether it stands to the left or to the right of box number 2. And if you are "taking into account the permutations" the answer is still $m^n$ - if you shuffle the boxes around it doesn't change the content of each box. $\endgroup$ Oct 1, 2018 at 9:10
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    $\begingroup$ Yes I think I got it now. And the order of the balls in a single box does not matter since a ball can just be in this box or not. There are no drawers in the boxes. And hence I am not able to take into account the order of the balls. Let's say B(1,2,3) is the same like B(2,1,3) since I can only say balls 1,2,3 are in this certain box. Is this correct? $\endgroup$
    – zodiac
    Oct 1, 2018 at 9:17
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    $\begingroup$ There is only interest in the end result of the distribution (which labeled balls have ended up in which labeled box?). Not in the way the distribution takes place. $\endgroup$
    – drhab
    Oct 1, 2018 at 9:17

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