Question

Calcule o valor dos seguintes números binomiais

Answer 1

Explicação passo-a-passo:

a) Pela relação de Stifel:

$\sf \dbinom{n}{p-1}+\dbinom{n}{p}=\dbinom{n+1}{p}$

$\sf \dbinom{7}{4}+\dbinom{7}{5}=\dbinom{8}{5}$

$\sf =\dfrac{8!}{5!\cdot(8-5)!}$

$\sf =\dfrac{8!}{5!\cdot3!}$

$\sf =\dfrac{8\cdot7\cdot6\cdot5!}{5!\cdot3!}$

$\sf =\dfrac{8\cdot7\cdot6}{3!}$

$\sf =\dfrac{336}{6}$

$\sf =56$

b)

$\sf \dbinom{6}{4}+\dbinom{6}{5}-\dbinom{7}{4}$

$\sf =\dbinom{7}{5}-\dbinom{7}{4}$

$\sf =\dfrac{7!}{5!\cdot(7-5)!}-\dfrac{7!}{4!\cdot(7-4)!}$

$\sf =\dfrac{7!}{5!\cdot2!}-\dfrac{7!}{4!\cdot3!}$

$\sf =\dfrac{7\cdot6\cdot5!}{5!\cdot2!}-\dfrac{7\cdot6\cdot5\cdot4!}{4!\cdot3!}$

$\sf =\dfrac{7\cdot6}{2!}-\dfrac{7\cdot6\cdot5}{3!}$

$\sf =\dfrac{42}{2}-\dfrac{210}{6}$

$\sf =21-35$

$\sf =-14$

c)

$\sf \dbinom{10}{4}+\dbinom{2}{0}+\dbinom{8}{3}$

$\sf =\dfrac{10!}{4!\cdot(10-4)!}+\dfrac{2!}{0!\cdot(2-0)!}+\dfrac{8!}{3!\cdot(8-3)!}$

$\sf =\dfrac{10!}{4!\cdot6!}+\dfrac{2!}{0!\cdot2!}+\dfrac{8!}{3!\cdot5!}$

$\sf =\dfrac{10\cdot9\cdot8\cdot7\cdot6!}{4!\cdot6!}+1+\dfrac{8\cdot7\cdot6\cdot5!}{3!\cdot5!}$

$\sf =\dfrac{10\cdot9\cdot8\cdot7}{4!}+1+\dfrac{8\cdot7\cdot6}{3!}$

$\sf =\dfrac{5040}{24}+1+\dfrac{336}{6}$

$\sf =210+1+56$

$\sf =267$

d)

$\sf \dbinom{6}{2}+\dbinom{12}{7}+\dbinom{7}{0}$

$\sf =\dfrac{6!}{2!\cdot(6-2)!}+\dfrac{12!}{7!\cdot(12-7)!}+\dfrac{7!}{0!\cdot(7-0)!}$

$\sf =\dfrac{6!}{2!\cdot4!}+\dfrac{12!}{7!\cdot5!}+\dfrac{7!}{0!\cdot7!}$

$\sf =\dfrac{6\cdot5\cdot4!}{2!\cdot4!}+\dfrac{12\cdot11\cdot10\cdot9\cdot8\cdot7!}{7!\cdot5!}+1$

$\sf =\dfrac{6\cdot5}{2!}+\dfrac{12\cdot11\cdot10\cdot9\cdot8}{5!}+1$

$\sf =\dfrac{30}{2}+\dfrac{95040}{120}+1$

$\sf =15+792+1$

$\sf =808$