Question

Combinari de n+4 luate cate n+3=5

Answer 1

$\text{Conditii de existenta:} n\geq -3 ,n\in \mathbb{Z}\\C_{n+4}^{n+3}= 5\\\dfrac{(n+4)!}{(n+3)!\cdot 1!}=5\\\dfrac{(n+3)!\cdot (n+4)}{(n+3)!}=5\\n+4=5\\\boxed{n=1}\\\text{Hope you understood.}$