Question

me ajuda ????? e urgente
1) simplifique as expressões
a)An,p/Cn,p
b)Cn,n-1

Answer 1

Olá.

Para simplificar essas expressões, devemos levar em consideração suas formas "extensas":

$\mathsf{A_{n,~p}=\dfrac{n!}{(n-p)!}}\\\\\\ \mathsf{C_{n,~p}=\dfrac{n!}{p!\cdot(n-p)!}}$

Iremos usar cálculos algébricos elementares. Vamos aos cálculos da primeira expressão.

$\mathsf{\dfrac{A_{n,p}}{C_{n,p}}=\dfrac{n!}{(n-p)!}\div\dfrac{n!}{p!\cdot(n-p)!}}\\\\\\ \mathsf{\dfrac{A_{n,p}}{C_{n,p}}=\dfrac{n!}{(n-p)!}\cdot\dfrac{p!\cdot(n-p)!}{n!}}\\\\\\ \mathsf{\dfrac{A_{n,p}}{C_{n,p}}=p!\cdot\dfrac{(n-p)!\cdot n!}{(n-p)!\cdot n!}}\\\\\\ \mathsf{\dfrac{A_{n,p}}{C_{n,p}}=p!}$

O cálculo da segunda será:

$\mathsf{C_{n,~n-1}=\dfrac{n!}{(n-1)!\cdot(n-(n-1))!}}\\\\\\ \mathsf{C_{n,~n-1}=\dfrac{n\cdot(n-1)!}{(n-1)!\cdot1!}}\\\\\\ \mathsf{C_{n,~n-1}=\dfrac{n}{1}=n}$

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Caso a interpretação seja C_{n, n} - 1, teremos:

$\mathsf{C_{n,~n}-1=\dfrac{n!}{n!\cdot(n-n)!}-1}\\\\\\ \mathsf{C_{n,~n}-1=\dfrac{n!}{n!\cdot0!}-1}\\\\\\ \mathsf{C_{n,~n}-1=\dfrac{1}{1\cdot1}-1}\\\\\\ \mathsf{C_{n,~n}-1=1-1=0}$

Qualquer dúvida, deixe nos comentários.

Bons estudos.