Is there an algorithm to find the lengths of all paths from a given vertice that can run in polynomial time? I have seen from a few sources (such as the CS Stack Exchange) that the problem of determining all the paths between two vertices $u$ and $v$ is NP hard. However, foR something that I am developing, I need to find only the number of paths of a given length whose first vertice is $v$. Is there an algorithm that can find this in Polynomial time ?
P.S. This is the first time I am posting on Stackexchange, so I apologize in case my question is too "rough" .

If you only need the number of walks of a given length, you can do it in polynomial time (with respect to the number of vertices of the graph) using the incidence (or adjacency) matrix of the graph.
Indeed, assume $G= (V,E)$ is the graph in question, where $|V| = n$ and $A$ is the incidence matrix of $G$, i.e. $A$ is an $n\times n$ matrix of $\{0,1\}$-s where $(i,j)$-th element is $1$ iff there is an edge from vertex $i$ to vertex $j$. 
Fix an integer $k \geq 1$, which is the length of the walk. It is a well-known fact that the $(i,j)$-th entry of the matrix $A^k$ shows the number of walks from $i$ to $j$ having length $k$ (the proof follows by a straightforward induction on $k$ and the definition of a matrix product, see wikipedia for instance).
Getting back to your original question: $A^k$ can be computed in $O(n^3 + n\log k)$ time using singular value decomposition (SVD). Here $n^3$ is the time to complete the SVD and $n \log k$ is the time to raise all elements on the diagonal matrix of the SVD into power $k$. See here for further discussion.
Once you have $A^k$, then summing over all elements of the $v$-th row ($v$ was the vertex where you start the walk of length $k$) gives the number of walks from $v$ having length $k$.
For the number of paths (instead of walks), see the other answer.
