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Can someone explain this one (both a and b... I can't get the algebra to work out)? | Can someone explain this one (both a and b... I can't get the algebra to work out)? | ||
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| + | I don't know how to explain that they are counting the same thing, but I can explain the algebra. | ||
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| + | First, looking at the left side you have C(2n,2) which equals (2n)!/(2!*(2n-2)!). This can be "simplified" to (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*(n!/(2!*(n-2)!))+n^2. Simplifing this leads to (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). | ||
Revision as of 17:20, 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 (2n)!/(2!*(2n-2)!). This can be "simplified" to (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*(n!/(2!*(n-2)!))+n^2. Simplifing this leads to (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).
