
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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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:
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 setsoftwo?] If the order does not matter, there is one way:[color=beige] .[/color] [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: 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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