Consider this expression:
$$(-1)^{G+T+1} 2^{G-2T-2} G (G-1)(G-2) \frac{(G-T-2)!}{T!(G-2T)!}$$
Here, $G,T\in \mathbb{Z}$.
If we set $G = 2T$, it becomes
$$(-1)^{T+1}(2T-1) .$$
If we now set $T=0$, this evaluates to one. So far so good. Note that since $G=2T$, this obviously corresponds to $G=0$.
But if I try to set $G=0$ first, the original expression evaluates to zero. To make it worse, if I first set $T=0$, the original expression becomes
$$(-1)^{G+1} 2^{G-2}(G-2)$$
and it evaluates to one half when setting $G=0$.
I understand that taking limits in different order cannot always be done and I understand that, technically speaking, expressions of type $n!/(n-1)!$ aren't always strictly equal to $n$. Could someone shed some more light on what exactly is going on here? I'm a big boy and I don't like being confused by elementary stuff :)
For the first expression (when you first plug in $G = 2T$), the problem is a domain issue with the simplification. Substituting $G=2T$, we have $$ (-1)^{G+T+1}2^{G-2T-2}G(G-1)(G-2)\frac{(G-T-2)!}{T!(G-2T)!} = (-1)^{3T+1}2^{-2}(2T)(2T-1)(2T-2)\frac{(T-2)!}{T!}. $$
However, as you've pointed out, this last fraction $\frac{(T-2)!}{T!}$ doesn't always simplify to $\frac{1}{T(T-1)}$. In fact, it doesn't even make sense when $T -2 < 0$, without extending the definition of $!$ with something like the gamma function.
Similarly, if you first set the value of $T$ to $0$, then it becomes $$ (-1)^{G+T+1}2^{G-2T-2}G(G-1)(G-2)\frac{(G-T-2)!}{T!(G-2T)!} = (-1)^{G+1}2^{G-2}G(G-1)(G-2)\frac{(G-2)!}{G!}, $$ and you run into the same problems.
Without further context on the purpose of this expression (if it's just for recreation, that's fine too!), there isn't a clear "best" way to resolve the domain issue when $G - T - 2 < 0$, $T < 0$, or $G - 2T < 0$. Here, the first is the one that is causing problems.