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Algebra Help (Read 570 times)

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hero_bash

#1 April 21, 2009, 06:46:46 am

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So I have been doing these papers for a few hours now(whole day). And I got stumped by I guess simple algebra stuff.
So my question is say for example I have:

the set:
_, _, _, O, O, X,

and I will compute for the possible combinations.. well, I don't really know how to compute this. For one, order is important, however I don't want the _s and Os to count as different from the other _s and Os. So it's kind of a cross between combination and permutation.

EXAMPLE:
if i solve this by combination:
OOX and OOX would be counted as 2 combinations because it identified the two Os as different from one another.
--EDIT
--or will it???
--

How to do this?

Last Edit: April 21, 2009, 06:56:23 am by Evil Pikachu

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Frisky SKeleton

#2 April 21, 2009, 11:09:07 am

aye ess dee eff el cay jay ache
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what? if the O's and the _'s are the same then you really only have three items, just change the number of possible selections or whatever

I USE Q'S INSTEQD OF Q'S

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Guana

#3 April 21, 2009, 11:37:49 am

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Do you have to use all objects in the permutations (for example _O_O_X)? Or are permutations like OOX and OXO_ allowed?

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hero_bash

#4 April 21, 2009, 11:42:43 am

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All of it must be included

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Guana

#5 April 21, 2009, 11:58:49 am

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Then there's a simple formula for this. Suppose you have a total of n objects and k different ones. Now let n_1 be the amount of type 1 objects, n_2 the amount of type 2 objects, etc. Then the amount of permutations is

n! / (n_1! * n_2! * .. * n_k!)

And in this case, (6!) / (3! * 2! * 1!) = 60 since there are three _'s, two O's and one X which makes a total of 6 objects.

Honestly I don't remember why this works so I dunno if you really learn anything from this post!

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JMickle

#6 April 21, 2009, 02:20:59 pm

I gots a lego!
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i remember working that out in the middle of a test because i forgot it. it would seriously have been quicker to have listed all the combinations and counted them.

www.jmickle.com

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