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ericttz
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Probably easy. I am just stumped at how to get there. 2 09 Jan 2007, 07:35
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Probably easy. I am just stumped at how to get there.

2 table sand 2 chairs are to be selected, all are different. There are five chairs and 150 combinations. How many tables are there?

4
6
10
24
30

I tried to take the answer and plug into perm. or comb. but it doesnt seem to work. Not sure how to proceed.

Thanks in advance
Eric



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mbagal1
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Probably easy. I am just stumped at how to get there. 2 09 Jan 2007, 08:03
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is it not 5C2*(XC2) = 150?
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Sumithra
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Probably easy. I am just stumped at how to get there. 2 09 Jan 2007, 08:09
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hsampath
is it not 5C2*(XC2) = 150?
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I came up with the same approach and got B as ans.
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subhen
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Probably easy. I am just stumped at how to get there. 2 09 Jan 2007, 08:24
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isn't it amazing that sometimes simple problem twisted slightly makes us think so hard.

5C2*XC2=150
XC2=15
X!/(X-2)!=30

From this one should be able to figure out that X is 6.
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ericttz
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Probably easy. I am just stumped at how to get there. 2 09 Jan 2007, 08:42
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So I guess you only need to figure out one side of the problem. That is what threw me off. So take 6!/(6-4)! =30 and multiply by 5. Thats now how I would back solve it. I was trying to solve both and multiply Silly me.

Thanks a lot for the help. Everyone we get our mind around is one more chance to get one right on the test.

Eric
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ericttz
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Probably easy. I am just stumped at how to get there. 2 09 Jan 2007, 19:45
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Havent been here enough. Can you explain the shorthand 5C2 and XC2?

Not combination...doesnt work

an explanation would be much appreciated

Thanks for all the help
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ericttz
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Probably easy. I am just stumped at how to get there. 2 09 Jan 2007, 19:52
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I figured it out. The C threw me. Its a perm. not comb.

Thanks
Eric
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Sumithra
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Probably easy. I am just stumped at how to get there. 2 09 Jan 2007, 20:12
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ericttz
Havent been here enough. Can you explain the shorthand 5C2 and XC2?

Not combination...doesnt work

an explanation would be much appreciated

Thanks for all the help
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It is combination. 5C2=5!/2!(5-2)!=10
5C2 * xC2 = 150
10*xC2=150
xC2=15

Note:(x*(x-1))/2 is the same as picking combinations of two from x or xC2)

so you can derive x as 6*5/2=15
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Andr359
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Probably easy. I am just stumped at how to get there. 2 09 Jan 2007, 20:56
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If we are dealing with combs:

5C2 * nC2 = 10 * nC2 = 150. Then, nC2 = 15 = nC2 = n*(n-1)/2, then n*(n-1) = 30 => n = 6.

If we are dealing with perms:

5P2 * nP2 = 20 * n*(n-1) = 150 => n*(n-1) = 15/2, doesn´t fit the requirement that n be integer.



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Hi there,
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