Question

(UFPR) Se An,p=30 e Cn,p=15, ache o valor de (n+p)!/n!

Answer 1

An,p =n!/(n-p)! = Cn,p= n!/p!(n-p)! =>  An,p/Cn,p= [n!/(n-p)!] /[n!/p!(n-p)!]
 An,p/Cn,p= [n!/(n-p)!] /[n!/p!(n-p)!]=30/15 =2  => p!=2=>p=2

An,2 =n!/(n-2)!= 30 => n(n-1)(n-2)!/(n-2)!=30  => n(n-1)=30=> n^2-n-30=0
 delta =121,         n= (1+11)/2 =12/6 =2 

então (6+2)!/6! =8!/6!=8.7.6!/6! =8.7 =56

Answer 2

 Olá Jeehsm! Boa questão essa, hein!
 
 Para resolvê-la, precisamos saber que $\mathsf{A_{n, p} = \frac{n!}{(n - p)!}}$ e $\mathsf{C_{n, p} = \frac{n!}{p!(n - p)!}}$.
 
 Com efeito,

$\\ \mathsf{30 = 2 \cdot 15} \\\\ \mathsf{A_{n, p} = 2 \cdot C_{n, p}} \\\\ \mathsf{\frac{n!}{(n - p)!} = 2 \cdot \frac{n!}{p!(n - p)!}} \\\\\\ \mathsf{\frac{\diagup \!\!\!\!\! n!}{\underbrace{\mathsf{(n - p)!}}_{\mathtt{cancel}}} = 2 \cdot \frac{\diagup \!\!\!\!\! n!}{p! \underbrace{\mathsf{(n - p)!}}_{\mathtt{cancel}}}} \\\\\\ \mathsf{1 = 2 \cdot \frac{1}{p!}}$

$\\ \mathsf{p! = 2} \\\\ \mathsf{p! = 2 \cdot 1} \\\\ \mathsf{p! = 2!} \\\\ \boxed{\mathsf{p = 2}}$
 
 Por conseguinte, encontramos o valor de "n", veja:

$\\ \mathsf{A_{n, p} = 30} \\\\ \mathsf{\frac{n!}{(n - p)!} = 30} \\\\\\ \mathsf{\frac{n!}{(n - 2)!} = 30} \\\\\\ \mathsf{\frac{n \cdot (n - 1) \cdot (n - 2)!}{(n - 2)!} = 30}$

$\\ \mathsf{n \cdot (n - 1) = 30} \\\\ \mathsf{n \cdot (n - 1) = 6 \cdot 5} \\\\ \boxed{\mathsf{n = 6}}$
 
 Por fim,

$\\ \large \mathsf{\frac{(n + p)!}{n!} = \frac{(6 + 2)!}{6!}} \\\\\\ \mathsf{\frac{(n + p)!}{n!} = \frac{8!}{6!}} \\\\\\ \mathsf{\frac{(n + p)!}{n!} = \frac{8 \cdot 7 \cdot 6!}{6!}} \\\\\\ \mathsf{\frac{(n + p)!}{n!} = 8 \cdot 7} \\\\\\ \boxed{\boxed{\mathsf{\frac{(n + p)!}{n!} = 56}}}$
 
 Espero ter ajudado!!