Need help setting up a math problem

lennon3

Member
Apr 11, 2006
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I'm looking over a practice test about to go take my test, and there is one question on the practice test that i have no clue how to set up.

"Each of 2 countries sends 6 delgates to a negotiating conference. A rectangular table is used with 6 chairs on each long side. If each country is assigned a long side of the table (operation 1), how many seating arrangements are possible?"

I'm pretty sure it has something to do with combinations, but still don't know what to do next. The answer is 1,036,800 too btw.

Thanks to anyone who understands this!
 

mred

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Oct 19, 2006
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6! * 6! * 2

Each country has 6! ways of sitting in 6 seats, so that's 6! * 6! for two countries sitting on specific sides of the table. Since there are additionally two different configurations for which country is on which side (country A side 1 country B side 2, country A side 2 country B side 1) you multiply that by 2.
 
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ISUAgronomist

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Nov 5, 2009
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On the farm, IA
First seat has 12 choices (all others on that side will be from the same country). The other side will be your standard 6!

12*5*4*3*2*1*6*5*4*3*2*1=1036800

EDIT: What mred said would be the correct answer. I missed the "assigned" sides keyword.
 
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ist8er

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Aug 3, 2010
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on one side : 6*5*4*3*2*1=720
on the other side : 6*5*4*3*2*1=720
both sides combined : 720*720=518,400
if the countries switched sides : 518,400*2=1,036,800
 

lennon3

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Apr 11, 2006
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I understand it now. Didn't really want to go into the lab for one quick question!! Thanks a lot for the help guys!
 

Chipper

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Dec 8, 2008
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on one side : 6*5*4*3*2*1=720
on the other side : 6*5*4*3*2*1=720
both sides combined : 720*720=518,400
if the countries switched sides : 518,400*2=1,036,800
 

CycloneGB

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Jul 20, 2010
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Cyforce

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Nov 24, 2009
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I came up with exactly half that number then it dawned on me.
The countries can switch sides and that gives you the answer.
 

lennon3

Member
Apr 11, 2006
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Okay got another one for you guys, this one is probably a really easy one but confusing to me. Its the only one I missed on the practice test so I just need to figure this one out.

Given two throws of a fair die, what is the probability you get at least a 4 on each throw? At least a 5 on the first or second throw?

I got the 4 part right with 1/4. But the missed the five one. I put 2/3 and the answer is 5/9.