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I don't know how to explain that they are counting the same thing, but I can explain the algebra. | I don't know how to explain that they are counting the same thing, but I can explain the algebra. | ||
| − | First, looking at the left side you have C(2n,2) which equals (2n)! | + | First, looking at the left side you have C(2n,2) which equals <math>\frac{(2n)!}{(2!*(2n-2)!}</math>. This can be "simplified" to <math>\frac{(2n*(2n-1))}{2}</math>. which further simplifies down to <math>2n^2-n</math>. The right side starts at <math>2*C(n,2)+n^2</math>. so <math>2*(\frac{n!}{(2!*(n-2)!)})+n^2</math>. Simplifing this leads to <math>\frac{(2*n*(n-1))}{2} + n^2</math> which goes to <math>n^2 - n + n^2</math> which ofcourse equals <math>2n^2 - n</math>. (someone that understands how to make the formulas look nice, feel free to do so). |
Revision as of 17:13, 17 September 2008
Can someone explain this one (both a and b... I can't get the algebra to work out)?
I don't know how to explain that they are counting the same thing, but I can explain the algebra.
First, looking at the left side you have C(2n,2) which equals $ \frac{(2n)!}{(2!*(2n-2)!} $. This can be "simplified" to $ \frac{(2n*(2n-1))}{2} $. which further simplifies down to $ 2n^2-n $. The right side starts at $ 2*C(n,2)+n^2 $. so $ 2*(\frac{n!}{(2!*(n-2)!)})+n^2 $. Simplifing this leads to $ \frac{(2*n*(n-1))}{2} + n^2 $ which goes to $ n^2 - n + n^2 $ which ofcourse equals $ 2n^2 - n $. (someone that understands how to make the formulas look nice, feel free to do so).
