Question

Calcule o valor das combinações simples:
A)C n,3- C n,2=0
B)C n+3 ,n+1=28

Answer 1

$\mathrm{\mathbf{01.}\ C_{n,3}-C_{n,2}=0\ \to\ C_{n,3}=C_{n,2}}\\\\ \mathrm{\dfrac{n!}{3!(n-3)!}=\dfrac{n!}{2!(n-2)!}\ \to\ \dfrac{1}{6(n-3)!}=\dfrac{1}{2(n-2)(n-3)!}}\\\\\\ \mathrm{\dfrac{1}{6}=\dfrac{1}{2(n-2)}\ \to\ 2n-4=6\ \to\ 2n=10\ \to\ \boxed{\mathbf{n=5}}}$

$\mathrm{\mathbf{02.}\ C_{n+3,n+1}=28\ \to\ \dfrac{(n+3)!}{(n+1)!(n+3-(n+1))!}=28}\\\\\\ \mathrm{\dfrac{(n+3)(n+2)(n+1)!}{(n+1)!(n+3-n-1)!}=28\ \to\ \dfrac{(n+3)(n+2)}{2!}=28}\\\\\\ \mathrm{(n+3)(n+2)=2.28\ \to\ n^2+2n+3n+6=56}\\\\ \mathrm{n^2+5n+6-56=0\ \to\ n^2+5n-50=0}\\\\ \mathrm{n=\dfrac{-5\pm\sqrt{5^2-4.1.(-50)}}{2.1}=\dfrac{-5\pm\sqrt{25+200}}{2}}\\\\ \mathrm{n=\dfrac{-5\pm\sqrt{225}}{2}=\dfrac{-5\pm15}{2}\ \to\ Como\ n\ \textgreater \ 0:}\\\\ \mathrm{n=\dfrac{-5+15}{2}=\dfrac{10}{2}\ \to\ \boxed{\mathbf{n=5}}}$