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Suppose I have $n$ objects(non-identical) and I want to distribute them in $r$ groups.Let's say $n=r=10$ .The solution given is $10^{10}$. I cannot understand this result at all. Well I know if I try to make cases it seems summing up all the $\prod \binom{n}{r}$ will give $(1+9)^{10}$ but in the book it's done as- ways for distributing objects for each box is $n$ and since there are $r$ boxes then total ways are $n^r$. Can someone help me understand this method.

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  • $\begingroup$ The answer $10^{10}$ is valid only if the groups are also distinct. $\endgroup$ – true blue anil Nov 30 '15 at 4:28
  • $\begingroup$ and if you can put an object in more than one group. $\endgroup$ – Christopher Carl Heckman Nov 30 '15 at 4:41
  • $\begingroup$ By "distinct", I meant distinguishable, or labelled $\endgroup$ – true blue anil Nov 30 '15 at 5:00
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The $r$ groups are labelled (though not explicitly stated)

Now each object has $r$ choices for group,

and we just apply the multiplication principle to get $r\cdot r \cdot r.. = r^n$

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  • $\begingroup$ What do you mean by "not distinct" groups? Is it such that one object can be placed in two or more groups? Doesn't quite make sense to me. $\endgroup$ – cr001 Nov 30 '15 at 4:52
  • $\begingroup$ By distinct, I mean labelled. Changing terminology in the answer. $\endgroup$ – true blue anil Nov 30 '15 at 4:57
  • $\begingroup$ Okay I got what you mean. For example if grouping (in terms of cardinality of each group) $(2,3,1)$ and $(1,3,2)$ are the same then indeed the answer does not work. $\endgroup$ – cr001 Nov 30 '15 at 5:00

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