[6447] in s-news-athena
Re: COMBINATIONAL PROBLEM 15/2/95
daemon@ATHENA.MIT.EDU (Francoise Gelis)
Wed Feb 15 06:11:33 1995
Date: Wed, 15 Feb 1995 11:36:26 GMT
From: Francoise Gelis <fg@baobab.jouy.inra.fr>
To: s-news@utstat.toronto.edu
The unique number of combinations whose elements do NOT share the same
set seems to be : n1 * n2 * ... * np, where (ni) are the numbers of
elements in each set, and p the number of sets. In the example, we
obtain 2 * 2 * 2 = 8.
Francoise Gelis
----------------------------------------------------------------------
INRA Francoise.Gelis@jouy.inra.fr
Laboratoire de Biometrie
78352 Jouy-en-Josas Cedex
France
----------------------------------------------------------------------
Lotto Player asked:
> I have 3 sets of combinations sets a,b,c
> Each set has the following numbers in it
> For simplicity, I have numbered the set elements sequentially
>
> set a 1,2
> set b 3,4
> set c 5,6
>
> This gives one set d as 1,2,3,4,5,6 or 6 elements
>
> I wish to calculate all combinations of the 3 sets
> However, I wish to eliminate those combinations which have
> elements which share the same set
>
> I assume we would use select 3 from 6 as a starting point, producing 20 combinations.
> However several of these combinations would have elements which share the same
> set. Those combinations which have an x marked next to them have elements which share
> the same set. There are 12 combinations having elements which share the same set. The
> total of 20 combinations, less the 12 which share the same set gives us 8 unique combination
> sets. This is what I wish to end up with.
>
> My question is as follows.
> Is there a formula which can be used to calculate the unique number of combinations
> whose elements do NOT share the same set as shown in this example ???
> And if so, cold you tell me what it is ?
> Thanks
> David
>
> COMBINATIONS ( 3 from 6 = 20)
> 1,2,3 x
> 1,2,4 x
> 1,2,5 x
> 1,2,6 x
> 1,3,4 x
> 1,3,5
> 1,3,6
> 1,4,5
> 1,4,6
> 1,5,6 x
> 2,3,4 x
> 2,3,5
> 2,3,6
> 2,4,5
> 2,4,6
> 2,5,6 x
> 3,4,5 x
> 3,4,6 x
> 3,5,6 x
> 4,5,6 x
