In how many ways can you pick $m$ cats from $n$ cats if $2$ given cats must never be together at the $m$-group? In how many ways can you pick $m$ cats from $n$ cats if $2$ given cats must never be together at the $m$-group?

Is it $C(n,m)-C(n-2,m-2)$,  $C(n-2,m)+C(n-1,m)+C(n-1,m)$ or something else?
The first answer is correct, the second is wrong.

The first formula amounts to (all combos) - (combos with both the cats together)

The second formula tries to compute (neither cat in combo) + (cat 1 in combo) + (cat 2 in combo), but fails to see that the 2nd and 3rd terms do not guarantee the specific cat actually being in the combo.

The correct expression would be $\dbinom{n-2}{m} + 2\dbinom11\dbinom{n-2}{m-1}=  \dbinom{n-2}{m} + 2\dbinom{n-2}{m-1}$
Both answers are true. To see the equality you can use $\dbinom{n}{k}=\dbinom{n-1}{k-1}+\dbinom{n-1}{k}$.