Re: INFORMATIONS
- To: mathgroup at smc.vnet.net
- Subject: [mg17346] Re: [mg17292] INFORMATIONS
- From: BobHanlon at aol.com
- Date: Mon, 3 May 1999 01:45:52 -0400
- Sender: owner-wri-mathgroup at wolfram.com
In a message dated 5/1/99 12:36:27 AM, lwpao at tin.it writes:
>I am a new "Mathematica" user.
>I'm not a mathematician but interested in "Mathematica" possibilities about
>combinatory calculations.
>I've read the about Help file but I haven't understood how in practice
>I can
>do it. How to order the computer to do for example this:
>input: 3 5 15 22 52 90
>output: show me all the possible combinations for those numbers.
>Can you teach me about, showing me step by step what I can do?
>My "Mathematica" version is 3.0 and my operative system is Mac OS 8.5
>
>I'd like also informations about Italian sites for "Mathematica" and about
>combinatorica related to Lottos and Lotteries (also in english).
>
Giuseppe,
selection = {3, 5, 15, 22, 52, 90};
n = Length[selection];
For n numbers, there are n! possible permutations (different orderings).
For n =6 that is 6! = 720, i.e.,
n! == 6! == 720
True
Loading the standard add-on package DiscreteMath`Combinatorica`
Needs["DiscreteMath`Combinatorica`"]
These n! permutations are given by the function MinimumChangePermutations
Length[MinimumChangePermutations[selection]]
720
There are n!/(m! * (n-m)!) subsets of these n objects taken m at a time.
For example, for n = 6 and m = 2 there are 6!/(2! * 4!) = 15 subsets
KSubsets[selection, 2]
Length[%]
{{3, 5}, {3, 15}, {3, 22}, {3, 52}, {3, 90}, {5, 15},
{5, 22}, {5, 52}, {5, 90}, {15, 22}, {15, 52}, {15, 90},
{22, 52}, {22, 90}, {52, 90}}
15
And @@ Table[Length[KSubsets[selection, m]] == n!/(m! * (n-m)!), {m, 0, n}]
True
For all possible values of m, there are 2^n such subsets (every possibility
where each element is either included or not)
Sum[k!/(m! * (k-m)!), {m, 0, k}] == 2^k // FullSimplify
True
where k was used instead of n since n was given the specific value of 6.
Using k shows that the result is true for all values of k including k = n = 6.
Bob Hanlon