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How many ways are there to distribute $15$ distinguishable objects into $5$ distinguishable boxes so that the boxes have one, two, three, four, and five objects in them ?


I tried to solve it as =>

$C(15,1) * C(14,2) * C(12,3) * C(9,4) * C(5,5)$ and

since placing total number of objects in each box matters here ,i.e, objects being placed in boxes in the count $1,2,3,4,5$ is different from $3,1,5,4,2$ and so on.

So, i multiplied it by $5!$


Hence, final answer I think should be

$5! * C(15,1) * C(14,2) * C(12,3) * C(9,4) * C(5,5)$


I don't have answer for this question. So, is my approach right ?

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Yes.   That looks alright.

You counted the ways to partition the objects into groups of distinct sizes $1,2,3,4,5$, and then ways to arrange those groups into the boxes.   That is what you wanted to count, and how you could count it.


Also written as $5!\dbinom{15}{5,4,3,2,1}$ using the multinomial coefficient notation.

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  • $\begingroup$ Thanks !! Can sterling formula be applied here ? $\endgroup$ – Jon Garrick Nov 16 '16 at 3:03
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    $\begingroup$ Stirling's formula states: $\lim\limits_{n\to\infty}\dfrac{n!}{n^{n+1/2}~e^{-n}}=\sqrt{~2~\pi~} $, which is not at all useful. You *may* be thinking of Stirling Numbers of the Second Kind $\left\{\begin{matrix}n\\k\end{matrix}\right\}$, which count ways to partition a set of $n$ elements into $k$ non-empty subsets. Though, this is not useful either. Or are you thinking of a different "Sterling" all together? $\endgroup$ – Graham Kemp Nov 16 '16 at 3:10
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    $\begingroup$ Yes, I was thinking of second kind. The season that we can't apply that here as size of boxes is fixed to contain objects, but sterling is based more on how to partition the objects in different ways . Am I right here? $\endgroup$ – Jon Garrick Nov 16 '16 at 3:14
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    $\begingroup$ That is so. Stirling's No2K counts ways to partition the set into a number of non-empty subsets of any size; whilst here we wish to count ways to partition into sets of given sizes. $\endgroup$ – Graham Kemp Nov 16 '16 at 3:30
  • $\begingroup$ Thanks again for clearing me :) $\endgroup$ – Jon Garrick Nov 16 '16 at 3:31

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