Hardy and wright define the following
$$\binom{m}{n} = \frac{m(m-1)(m-2)..(m-n+1)}{n!} \; ... (1)$$ $$\binom{-m}{n} = (-1)^n \;\frac{m(m+1)(m+2)..(m+n-1)}{n!} \; ...(2)$$
then they say $$\binom{-m}{n} = (-1)^n\binom{m+n-1}{n} \; ...(3)$$
Eq. $(3)$ does not make sense to me. How is it obtained ? ( they don't give any constraint on $m,n$, as to which is bigger or smaller ).
I think that you originally stated $(2)$ and $(3)$ incorrectly. Assuming that you’re referring to their Theorem $73$, they should read
$$\binom{-m}n=(-1)^n\frac{m(m+1)(m+2)\ldots(m+n-1)}{n!}\tag{2a}$$
and
$$\binom{-m}n=(-1)^n\binom{m+n-1}{n}\;,\tag{3a}$$
respectively (as they now do). Observe that $(1)$ says that
$$\binom{m+n-1}{n}=\frac{(m+n-1)(m+n-2)\ldots(m)}{n!}\;:$$
there are $n$ factors in the numerator, counting down from $m+n-1$ to $m$. These are the same $n$ factors that appear in the numerator of $(2\text{a})$ in the opposite order, so $3\text{a}$ is just a restatement of $2\text{a}$.
