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 > > > > > > >