Alice (A) and Bob (B) repeatedly play a game. A wins 4 times and B wins 3 times. In how many ways can you arrange the outcome of the games so that at some point A and B were tied?
Visualize a 5 by 4 grid on which the coordinates bijectively map to the possible scores. Then count all of the paths that cross through (1, 1), (2, 2), (3, 3).
$A_1 = \{\text{Paths that cross through } (1,1)\}$
$A_2 = \{\text{Paths that cross through } (2,2)\}$
$A_3 = \{\text{Paths that cross through } (3,3)\}$
we get $$\mid A_1 \cup A_2 \cup A_3\mid = \mid A_1\mid + \mid A_2\mid + \mid A_3\mid - \mid A_1 \cap A_2 \mid - \mid A_1 \cap A_3 \mid - \mid A_2 \cap A_3 \mid $$ $$ + \mid A_1 \cap A_2 \cap A_3 \mid = $$
$$ {2 \choose 1}{5\choose 3} + {4\choose 2}{3\choose 2} + {6\choose 3}{1 \choose 1} - {2\choose1}{2\choose1}{3\choose1} - {2\choose1}{4\choose2}{1\choose1} - {4\choose2}{2\choose1}{1\choose1} + {2\choose1}{2\choose1}{2\choose1}{1\choose1}$$
$$ = 30$$
Does this work?
Your method works perfectly fine. Due to the feature of the problem, I can give a little bit different solution. Consider a rectangular grid $(0,0),(4,0),(0,3),(4,3)$. Put red points on $(1,1),(2,2),(3,3)$. We construct a path from $(0,0)$ to $(4,4)$ using only positive $x$ and $y$ direction and each time we move a unit. With no restriction, there are $\binom 73=35$ ways to do it.
Now we calculate when will we $\textbf{not}$ touching the three red points. If the first step is choosen from $(0,0)$ to $(0,1)$, then at some point we must hit one of the red points, so we should go from $(0,0)$ straight to $(2,0)$. There are only five paths continuing from $(2,0)$ and not touching any of the red points: \begin{align*} (2,0)\to (2,1)\to (3,1)\to (3,2)\to (4,2)\to (4,3)\\ (2,0)\to (2,1)\to (4,1)\to (4,3)\\ (2,0)\to (3,0)\to (3,2)\to (4,2)\to (4,3)\\ (2,0)\to (3,0)\to (3,1)\to (4,1)\to (4,3)\\ (2,0)\to (4,0)\to (4,3) \end{align*} so yes, the answer is $35-5=30$. But your answer is the standard one.