GMATPrep question 24

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GMATPrep question 24

by mchen01 » Mon Nov 26, 2007 9:04 pm
Interesting combination problem!
Thanks for the help!

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by syv11 » Mon Nov 26, 2007 9:35 pm
6!/(4!2!)*n!/[(n-2)!2!]=150
n!/(n-2)!=30

Insert answer solutions for n

if you use 6 you get to 15

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by mchen01 » Tue Nov 27, 2007 6:31 am
Yup, the answer is 6, thanks!

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DON'T UNDERSTAND

by dferm » Wed Nov 28, 2007 10:42 am
I don't quite get your explanation for this problem. Can you please elaborate further?

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by aimhigh715 » Wed Nov 28, 2007 3:25 pm
The idea is total combination = combination of chairs x combination of tables.

Since we know it's 150 total combos possible that MUST mean : combo of chairs x combo tables

Chair combo is known to be : 5!/2!3! = 5 x 2 = 10

Thus combo of tables = 15 = x! / 2!(x-2)!
Work it backwards and you'll find that the only way to get 30 = x(x-1) is if x = 6

Hope that helped!
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