Answer to Question #274143 in Discrete Mathematics for Mark

Question #274143

Show that if n is a positive integer with n ≥ 3, then




C(n, n − 2) = ((3n − 1)C(n, 3))/4





1
Expert's answer
2021-12-23T18:01:12-0500
c(n,n2)=2(n3)+3(n4)c(n, n-2)=2\dbinom{n}{3}+3\dbinom{n}{4}

=2n!3!(n3)!+3n!4!(n4)!=2\dfrac{n!}{3!(n-3)!}+3\dfrac{n!}{4!(n-4)!}

=n(n1)(n2)3+n(n1)(n2)(n3)8=\dfrac{n(n-1)(n-2)}{3}+\dfrac{n(n-1)(n-2)(n-3)}{8}

=n33n2+2n3+n483n34+11n283n4=\dfrac{n^3}{3}-n^2+\dfrac{2n}{3}+\dfrac{n^4}{8}-\dfrac{3n^3}{4}+\dfrac{11n^2}{8}-\dfrac{3n}{4}

=n485n312+3n28n12=\dfrac{n^4}{8}-\dfrac{5n^3}{12}+\dfrac{3n^2}{8}-\dfrac{n}{12}

(3n1)C(n,3)4=(3n1)n!3!(n3)!(4)\dfrac{(3n-1)C(n,3)}{4}=\dfrac{(3n-1)n!}{3!(n-3)!(4)}

=(3n1)(n)(n1)(n2)24=\dfrac{(3n-1)(n)(n-1)(n-2)}{24}

=n485n312+3n28n12=\dfrac{n^4}{8}-\dfrac{5n^3}{12}+\dfrac{3n^2}{8}-\dfrac{n}{12}

Hence


c(n,n2)=(3n1)C(n,3)4c(n, n-2)=\dfrac{(3n-1)C(n,3)}{4}


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