Show that if n is a positive integer with n ≥ 3, then C(n, n − 2) = ((3n − 1)C(n, 3))/4

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$c (n, n - 2) = 2 (3 n) + 3 (4 n)$

$=2\dfrac{n!}{3!(n-3)!}+3\dfrac{n!}{4!(n-4)!}$

$=\dfrac{n(n-1)(n-2)}{3}+\dfrac{n(n-1)(n-2)(n-3)}{8}$

$=\dfrac{n^3}{3}-n^2+\dfrac{2n}{3}+\dfrac{n^4}{8}-\dfrac{3n^3}{4}+\dfrac{11n^2}{8}-\dfrac{3n}{4}$

$=\dfrac{n^4}{8}-\dfrac{5n^3}{12}+\dfrac{3n^2}{8}-\dfrac{n}{12}$

$\dfrac{(3n-1)C(n,3)}{4}=\dfrac{(3n-1)n!}{3!(n-3)!(4)}$

$=\dfrac{(3n-1)(n)(n-1)(n-2)}{24}$

$=\dfrac{n^4}{8}-\dfrac{5n^3}{12}+\dfrac{3n^2}{8}-\dfrac{n}{12}$

Hence

$c(n, n-2)=\dfrac{(3n-1)C(n,3)}{4}$