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some math help please...?

3 replies [Last post]
Sun, 02/12/2006 - 15:23
Nazhuret
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Joined: 12/31/1969

I'm worse than an earthworm when it comes to math (they know how to divide i hear...) so if y'all could help me out with this problem that would be great. Or at least point me in the right direction.

In a given set there are 20 unique items

How many combinations of exactly 4 of these items can be generated?

The relative positions of each of the elements DOES matter.

IOW: The sets (2, 18, 10, 4) and (18, 4, 2, 10) are fundamentally different for my purposes.

Anyway, I don't expect someone else to do the work for me (though if you would like that would be great..) so if anyone can just give me a formula or point me somewhere to find one.

Thanks

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Sun, 02/12/2006 - 16:03
#1
Sebastian
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Joined: 07/27/2008
Re: some math help please...?

Nazhuret wrote:
I'm worse than an earthworm when it comes to math (they know how to divide i hear...) so if y'all could help me out with this problem that would be great. Or at least point me in the right direction.

In a given set there are 20 unique items

How many combinations of exactly 4 of these items can be generated?

The relative positions of each of the elements DOES matter.

IOW: The sets (2, 18, 10, 4) and (18, 4, 2, 10) are fundamentally different for my purposes.

Anyway, I don't expect someone else to do the work for me (though if you would like that would be great..) so if anyone can just give me a formula or point me somewhere to find one.

Thanks

20 x 19 x 18 x 17 = 116,280

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Sun, 02/12/2006 - 17:24
#2
OutsideLime
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Joined: 12/31/1969
some math help please...?

And to break it down to make it make sense:

You have 20 unique items, and 4 positions. (A, B, C, and D.... the order of the positions does not matter at all, but we'll label them this way for explanation purposes.)

A can be any of the 20

B has only 19 options, since 1 of the original 20 has been used up in the A position.

C has only 18 options, since 2 of the original 20 have been used up in the A and B positions.

D has only 17 options, since 3 of the original 20 have been used up in the A, B, and C positions.

Multiply those possibilities together to get Sebastian's result: 116,280

It's a bit of math that is nicely solvable by applying logic... i like it when that happens...

~Josh

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Mon, 02/13/2006 - 01:38
#3
Nazhuret
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Joined: 12/31/1969
some math help please...?

Thanks guys. That was much more simple than I was thinking it was going to be with a smaller number than I had thought.

It's still friggin huge when dealing with cards...

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