Hypergeometric distribution, $Hg(1,a,b)$ follows Bernoulli with $Be(\frac{a}{a+b})$

Question

The probability function of Hypergeometric distribution , $Hg(n,a,b)$ is $$P(X=m)=\frac{\binom{a}{m}\binom{b}{n-m}}{\binom{a+b}{n}}$$

I have to show $Hg(1,a,b)$ follows Bernoulli with $Be(\frac{a}{a+b})$

I noticed if i put $m=1$ when $n=1$ then i get the desired $Be(\frac{a}{a+b})$.

But i have not understood why will not i consider $m=0$ ?