My Math Forum permutation/combination (?)

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 November 6th, 2011, 07:42 AM #1 Newbie   Joined: Jan 2010 Posts: 15 Thanks: 0 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
November 6th, 2011, 08:16 AM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

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]$\{A,\,B\},\;\{C,\,D\},\;\{E,\,F\},\;\{G,\,H\}$

So we have eight items to arrange in a row:[color=beige] .[/color]$A,\:B,\:C,\:D,\:E,\:F,\:G,\:H$
[So why were they grouped into sets-of-two?]

If the order does not matter, there is one way:[color=beige] .[/color]$A\:B\:C\:D\:E\:F\:G\:H$

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

 November 6th, 2011, 08:45 AM #3 Newbie   Joined: Jan 2010 Posts: 15 Thanks: 0 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
 November 6th, 2011, 08:55 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: permutation/combination (?) For this, we would use: $N={8 \choose 4}=70$ 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: $8\cdot7\cdot6\cdot5=1680$ arrangements. But, since order does not matter, we divide by the number of ways to arrange 4 objects: $4!=24$ $\frac{8\cdot7\cdot6\cdot5}{4\cdot3\cdot2\cdot1}=7\ cdot2\cdot5=70$

### i suck at combination maths

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