My Math Forum Combination Questions

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 October 5th, 2009, 01:58 PM #1 Newbie   Joined: Oct 2009 Posts: 2 Thanks: 0 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. Thanks.
 October 5th, 2009, 02:50 PM #2 Senior Member   Joined: Mar 2007 Posts: 428 Thanks: 0 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.
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??

 October 5th, 2009, 06:12 PM #4 Senior Member   Joined: Mar 2007 Posts: 428 Thanks: 0 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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# nC(n-2)=10

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