Re: The number of solutions to n_1 + n_2 + n_3 + ... + n_k = m
- To: mathgroup at smc.vnet.net
- Subject: [mg38305] Re: [mg38267] The number of solutions to n_1 + n_2 + n_3 + ... + n_k = m
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Thu, 12 Dec 2002 01:31:42 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
I don't think there is a "simple expression" in a conventional sense,
but you can give a simple "formula" in terms of a generating power
series:
f[k_, m_] := SeriesCoefficient[Series[1/(1 - x)^k, {x, 0, m - k}], m -
k]
Andrzej Kozlowski
Yokohama, Japan
http://www.mimuw.edu.pl/~akoz/
http://platon.c.u-tokyo.ac.jp/andrzej/
On Tuesday, December 10, 2002, at 06:11 PM, Kumar Chellapilla wrote:
> Hi,
>
> I was wondering if there was a simple expression (as a function of m
> and k )
> to compute
> the total number of solutions to the "integer programming" problem:
>
> n_1 + n_2 + n_3 + ... + n_k = m
>
> where m > k, each ni >= 1.
>
> Basically, I am trying to figure out the number of ways of partitioning
> a set with m unique elements into k non-empty disjoint subsets.
>
> For example, if k = 2, then the number of solutions to
>
> n_1 + n_2 = m is (m-1) = M_2(m) (say)
>
> if k = 3, then the number of solutions to
>
> n_1 + n_2 + n_3 = m is M_3(m) = sum_{i=1}^{m-2} (m-i-1) =
> sum_{i=1}^{m-2} M_2(m-i)
>
> I can think of a recursive solutions wherein
>
> M_2(m) = (m-1) and
> M_k(m) = sum_{i=1}^{m-k+1} M_(k-1)(m-i), k > 2
>
> Thanks,
> Kumar
>
>
>
>
>
>
>