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In how many ways to ordered partition an natural number $n$ so none of the parts is greater then $k$? Ordered Partitions $3=2+1$ is NOT equal to $3=1+2$,

Some examples:

Ex1: $n=3,k=2$, this leaves only $3=2+1, 3=1+2$ and $3=1+1+1$

Total Partitions are $1,1,1\color{red}{;}2,1\color{red}{;}1,2\color{red}{;}3$ so we have $4$ total and only $3$ are wanted.

Ex2: $n=4,k=2$ this leaves only $1,1,1,1\color{red}{;}2,1,1\color{red}{;}1,2,1\color{red}{;}1,1,2\color{red}{;}2,2$

total: $1,1,1,1\color{red}{;}2,1,1\color{red}{;}1,2,1\color{red}{;}1,1,2\color{red}{;}2,2\color{red}{;}1,3\color{red}{;}3,1\color{red}{;}4$

Its connected to this Problem: In how many ways can 40 identical carrots be distributed among 8 different rabbits?

There is a "solution". although not useable, at least for me.

Donald Splutterwit
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thetha
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  • Mathjax, please...https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Franklin Pezzuti Dyer Jul 24 '17 at 18:48
  • what happens if the imposed condition of max 16 carrots is removed? Is there a book on this? or other conditions imposed? Is there something for undergraduate. – thetha Jul 24 '17 at 20:55

1 Answers1

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First, we can work with ordered partitions in $k$ parts with size limit $w$. This is explained in Stars and bars with restriction of size between bars via generating functions.: you have $k$ parts, each which may have $1, \dots, w$ which is represented by the generating function $x+x^2+\dots+x^w = \frac{x(1-x^w)}{1-x}$. With $k$ piles, you find that the number of way to generate $n$ is the $n$th coefficient of $f(x) = \frac{x^k(1-x^w)^k}{(1-x)^k}$. This is \begin{align*} [x^n]f(x) &= [x^{n-k}]\frac{(1-x^w)^k}{(1-x)^k} = [x^{n-k}]\sum_r(-1)^rx^{wr}\binom{k}{r}\sum_s\binom{s+k-1}{s}x^s \\ &= \sum_{wr+s=n-k} (-1)^r\binom{k}{r}\binom{s+k-1}{k-1} = \sum_r (-1)^r\binom{k}{r}\binom{n-wr-1}{k-1}. \end{align*} Now, just sum over $k$ to allow for any number of parts, so your desired answer is $$ \sum_k\sum_r (-1)^r\binom{k}{r}\binom{n-wr-1}{k-1}. $$ Hope that hideous sum helps! I'm not sure if there's a simpler form, I'd love for someone else to verify/correct me.

Tai
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