Question

3)Utilizando os casos especiais determine o valor de: (Com cálculo se possível)

Answer 1

Explicação passo-a-passo:

De modo geral:

• $\sf \dbinom{n}{n}=\dbinom{n}{0}=1$

• $\sf \dbinom{n}{1}=\dbinom{n}{n-1}=n$

• $\sf \dbinom{n}{k}=\dfrac{n!}{k!\cdot(n-k)!}$

a) $\sf \dbinom{7}{7}=1$, pois:

$\sf \dbinom{7}{7}=\dfrac{7!}{7!\cdot(7-7)!}$

$\sf \dbinom{7}{7}=\dfrac{7!}{7!\cdot0!}$

$\sf \dbinom{7}{7}=1$

b) $\sf \dbinom{5}{0}=1$, pois:

$\sf \dbinom{5}{0}=\dfrac{5!}{0!\cdot(5-0)!}$

$\sf \dbinom{5}{0}=\dfrac{5!}{0!\cdot5!}$

$\sf \dbinom{5}{0}=1$

c) $\sf \dbinom{27}{1}=27$, pois:

$\sf \dbinom{27}{1}=\dfrac{27!}{1!\cdot(27-1)!}$

$\sf \dbinom{27}{1}=\dfrac{27!}{1!\cdot26!}$

$\sf \dbinom{27}{1}=\dfrac{27\cdot26!}{26!}$

$\sf \dbinom{27}{1}=27$

d) $\sf \dbinom{30}{29}=30$, pois:

$\sf \dbinom{30}{29}=\dfrac{30!}{29!\cdot(30-29)!}$

$\sf \dbinom{30}{29}=\dfrac{30!}{29!\cdot1!}$

$\sf \dbinom{30}{29}=\dfrac{30\cdot29!}{29!}$

$\sf \dbinom{30}{29}=30$