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November 6th, 2011, 07:42 AM   #1
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permutation/combination (?)

hi
i hope this is the right board to post in.
i suck at math.

anyway, i have 4 groups with two items each.
how can i figure out in how many ways i can display all 8 items? the order does not matter
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November 6th, 2011, 08:16 AM   #2
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Re: permutation/combination (?)

Hello, mathtarded!

If you stated the problem correctly, you really do suck at math!


Quote:
I have 4 groups with two items each.
How can i figure out in how many ways i can display all 8 items?
The order does not matter.

The problem is rather silly, isn't it? . . . I might say STUPID.


We have four sets with two items each:[color=beige] .[/color]

So we have eight items to arrange in a row:[color=beige] .[/color]
[So why were they grouped into sets-of-two?]


If the order does not matter, there is one way:[color=beige] .[/color]

[color=beige]. . [/color]Well, duh!

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November 6th, 2011, 08:45 AM   #3
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Re: permutation/combination (?)

lol

i guess i didnt post the problem correctly.

what i mean is

how many arrangements, like this:

a, c, e, g
a, c, e, h
a, c, f, g
a, c, f, h
a, d, e, g
a, d, e, h
a, d, f, g
a, d, f, h
b, c, e, g
b, c, e, h
b, c, f, g
b, c, f, h
b, d, e, g
b, d, e, h
b, d, f, g
b, d, f, h

okay, 16...back to elementary school for me. lol
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November 6th, 2011, 08:55 AM   #4
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Re: permutation/combination (?)

For this, we would use:



There are 8 choices for the first position, 7 for the second, 6 for the third and 5 for the fourth. If order mattered, we would have:



arrangements. But, since order does not matter, we divide by the number of ways to arrange 4 objects:



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