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I have the following problem solved, but the answer seems wrong to me.

Problem: How many subgraphs does $K_3$ have? Same question for $K_n$

Answer:

We will classify the subgraphs by the size of their vertex-sets. By definition, a graph may not be empty. There are $\left(\begin{array}{c} n\\ k\\ \end{array}\right)$ choices of vertex set of size k, and for each such choice there are $\left(\begin{array}{c} k\\ 2\\ \end{array}\right)$ edges of $K_n$ available for our choice as edges of the subgraph. There are $2^{\left(\begin{array}{c} k\\ 2\\ \end{array}\right)}$ different choices of edge set. Therefore, the answer is $$\sum_{k=1}^{n}\left(\begin{array}{c} n\\ k\\ \end{array}\right)2^{\left(\begin{array}{c} k\\ 2\\ \end{array}\right)}$$

What I don't understand: $\left(\begin{array}{c} k\\ 2\\ \end{array}\right)=\infty$ for $k=1$, so how is this possible? Is the answer right? My professor might have made a mistake...

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We know that if we are choosing $k$ vertices on $K_{n}$ graph, those $k$ vertices must form a $K_{k}$. The number of edges of a $K_{m}$ graph is equal to $(m-1)+(m-2)+(m-3)+...+1$. This is because if we look at a vertex in $K_{m}$, it is connected to $(m-1)$ vertices. The next vertex is connected to $(m-2)$ because we already counted one of them, and so on.

However, what interesting is that that sum is equal to ${m \choose 2}$. You can see why at this math stack exchange answer.

Thus, we can use ${k \choose 2}$ for the number of edges in $K_{k}$.

The mistake here is yours, though it's minor: ${1 \choose 2}=0$, not $\infty$! ${n \choose k}$ is treated as 0 when $k > n$. This makes sense with the problem, because there are no edges to connect when $k=1$.

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