Is there a finite number to the possible state of electron in an atom?
For degenerate eigenstates any (normalized) linear combination of them will be again an eigenstate of same energy. This in fact means that there are infinitely many of them. 

So where does the 5 (in case of $d$-orbitals) come from, when we actually have infinitely many? It means we can select at most 5 of them, which are orthogonal (or more generally speaking: linear independent). We are free to choose which 5, and thus there are (infinitely) many set of 5 $d$-orbitals. 

Here are 2 possible sets (again for the example of $d$-orbitals) that are typically used:


The general solutions are $Y_l^m(\theta,\phi)=f_l^m(\theta) g_m(\phi)$ for $l=2$ and $m=\{-2,-1,0,1,2\}$. Where we have
\begin{align}
f_l^m(\theta) &amp;= (-1)^m \sqrt{\frac{(2l+1)(l-m)!}{4\pi (l+m)!}} P_l^m(\cos \theta) \\
 g_m(\phi) &amp;= e^{i m \phi}
\end{align}
and $P_l^m(\cos \theta)$ are the associated Legendre polynomials. As those are complex functions due to $g_m(\phi) = e^{i m \phi}$ (except for $m=0$), those are hard to plot or imagine.
The real linear combinations from the above:
\begin{align}
    d_{z^2} &amp;= Y_2^0(\theta,\phi) \\
    d_{x^2-y^2} &amp;= \frac{1}{\sqrt2}\left[Y_2^{+2}(\theta,\phi) + Y_2^{-2}(\theta,\phi)\right] \\
    d_{xy} &amp;= \frac{1}{\sqrt2 i}\left[Y_2^{+2}(\theta,\phi) - Y_2^{-2}(\theta,\phi)\right] \\
    d_{xz} &amp;= \frac{1}{\sqrt2}\left[Y_2^{+1}(\theta,\phi) + Y_2^{-1}(\theta,\phi)\right]\\
    d_{yz} &amp;= \frac{1}{\sqrt2 i}\left[Y_2^{+1}(\theta,\phi) - Y_2^{-1}(\theta,\phi)\right]\\
\end{align}