I have to prove some result. And for that, I need to prove this new problem. To prove, $c_{1}\Gamma(z+b_{1})+c_{2}\Gamma(z+b_{2})+\ldots+c_{n}\Gamma(z+b_{n})=0$ has at most $(n-1)$ real positive solutions, where $c_{i}$ are arbitrary non zero real numbers and $b_{i}$ are arbitrary positive numbers, s.t. $b_{i+1}>b_{i}>0$ for all $i$ and $\Gamma(z)$ is gamma function.