Homework Help Binomial coefficients - 2 different applications - explanation please? en>fr fr>en By Volans Comments: 492, member since Wed Feb 07, 2007On Thu Jan 19, 2012 03:32 PM
Is anyone able to explain a binomial coefficient formula application to me? I just don't understand the difference in wikipedia's doughnut analogy here:
en.wikipedia.org . . .
Compared to a standard combination, like this one:
en.wikipedia.org . . .
It still seems like choosing a combination of subsets from a set. 3 Replies to Binomial coefficients - 2 different applications - explanation please? |
re: Binomial coefficients - 2 different applications - explanation please? en>fr fr>en By Mendel  Comments: 1788, member since Wed Feb 23, 2005On Thu Jan 19, 2012 08:42 PM
I'm going to preface this with the caveat that it's been ages since I've looked at math like this, so take everything I say with a grain of salt.
The difference that I see between the "donut" example and the standard combination is that I think the "donut" case allows you to choose the same item twice. If you had exactly 10 donuts and were choosing 3, then it would be a standard combination which would be (10!)/[3!(10-3)!], which equals 120. But the donut example gives a grand total of 220. So I think in that case there are 10 *types* of donuts but you can choose the same type multiple times (like 2 chocolate donuts).
Does that sound about right? Please post if/ when you get to the bottom of this, I'd really like to know what's going on here! |
re: Binomial coefficients - 2 different applications - explanation please? en>fr fr>en By Volans Comments: 492, member since Wed Feb 07, 2007On Sun Jan 22, 2012 02:35 PM
OK, I revisited this issue after a few days, and yes, you're right, a "multicombination" seems to describe a combination of "types" of elements of a set - i.e. you can choose the same element more than once.
This example algorithm explains it pretty well:
www.martinbroadhurst.com . . .
And so, to take account of the "combinations within the subset" - i.e. all the different ways you can combine multiple "n" elements of the same type of a certain subset size "k", you must multiply those k-1 multicombinations by the normal n-combination, i.e.
becomes
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re: Binomial coefficients - 2 different applications - explanation please? en>fr fr>en By mirrim   Comments: 723, member since Sun Apr 06, 2008On Mon Jan 23, 2012 06:16 AM
I am such a nerd but this thread makes me miss math 
I used to be so good at statistics and probability in school. Now that I am working, I still use arithmetic, dilutions and use the occasional formula, but that is it. This stuff just goes right over my head now  |