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October 5th, 2009, 01:58 PM   #1
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Combination Questions

1. Solve the equation for n algebraically.

nCn-2 = 10

Would I have to divide the left side by (n)(n)(n-1)(n-2), which would cancel out (n) and (n-2). Im stuck with what to do after that.


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October 5th, 2009, 02:50 PM   #2
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Re: Combination Questions

1. n!/[2!(n-2)!] = n(n-1)/2 ...

2. Figure it for case one, being 6 pens and 8 pencils, then for case two, being 7 pens and 7 pencils and compare the two results.
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October 5th, 2009, 03:00 PM   #3
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Re: Combination Questions

Quote:
Originally Posted by David
1. n!/[2!(n-2)!] = n(n-1)/2 ...

2. Figure it for case one, being 6 pens and 8 pencils, then for case two, being 7 pens and 7 pencils and compare the two results.
Ya, I solved #2.

So number 1 would be

nCn-2 = 10

nCn-2/(n)(n-1) = 10n(n-1)/2??
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October 5th, 2009, 06:12 PM   #4
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Re: Combination Questions

Try again, using the equality I gave you. I used the fact that nCn-2 = n!/2!(n-2)!, that is all. Now simplify to what I gave you and sub that into the original equation. If you are doing combinations, you should be past simple algebraic equations and so be able to finish it now.
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