Question

(n+3)!. (n-2)!/n!(n-1)!​

Answer 1

Resposta:

Se for (n+3). (n-2)! / [n! (n-1)!]

(n+3)*(n-2)(n-1)!/[n!*(n-1)!]

(n+3)*(n-2)/n!   fica assim