Solve for n:(n-2) + (n-4) + (n-6) = 319 2 2 2 I believe this is an n choose r pr

Question

Solve for n:(n-2) + (n-4) + (n-6) = 319
2 2 2

I believe this is an n choose r problem, at least I think it is. I'm sorry i wasn't able to be
parenthesis aroudn the 2's, there supposed to be directly under it. Thanks.

Explanation / Answer

[ (n-2)C2 ] + [ (n-4)C2 ] + [ (n-6)C2 ] = 319 =>[ (n-2)(n-3) ]/2 + [ (n-4)(n-5) ]/2 + [ (n-6)(n-7)]/2 = 319 =>(n^2 -3n+2)/2 + (n^2 -7n +12)/2 + (n^2 -11n +30)/2 = 319 =>(3n^2 -21n +44)/2 = 319 =>3n^2 -21n +44 = 638 =>3n^2 -21n -594 = 0 =>n^2 -7n -198 = 0 =>n^2 -18n +11n -198 = 0 =>(n + 11) (n - 18) = 0 n = -11, 18 As n>0 => n = 18

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