is $\frac{Γ(r+x)}{Γ(r)Γ(x+1) }$ the same as ${r+x-1 \choose x}$?

Question

is $\frac{Γ(r+x)}{Γ(r)Γ(x+1) }$ the same as ${r+x-1 \choose x}$ ?

I have two pmfs $$P(X=x\mid\theta)={r+x-1 \choose x}(1-\theta)^x\theta^r $$ and $$P(X=x\mid\theta)=\frac{Γ(r+x)}{Γ(r)Γ(x+1) }(1-\theta)^x\theta^r $$

I was wondering if they would yield the same MlE

Answer 1

Assuming that both $x$ and $r$ are integers, yes. Since $\Gamma(n)=(n-1)!$ for all positive integers, then $$ \frac{\Gamma(r+x)}{\Gamma(r)\Gamma(x+1)} = \frac{(r+x-1)!}{(r-1)!x!} = \binom{r+x-1}{x}. $$