Let A be the first player and B be the second player.

a. If a set $E$ with the properties described in the text (called *perfect matching*) exists, then whenever A removes a vertex, B can simply remove the unique vertex in $G$ connected to the just remover vertex by an edge in $E$. This ensures that B can always respond to any move done by A, and thus has a winning strategy.

**b. **The existence of a perfect matching for a graph implies that the graph has an even number of vertices (every edge in $E$ connects two vertices, and touches every vertex exactly once). But $G$ has $15$ vertices and so no perfect matching exists.

**c. **See attached picture. Playing this way, there is only one moment in which B has a choice (on their second move), and they lose in both cases.