The number of paths on a graph of a fixed length w/o repeatings Sorry for bad English.
Consider a graph $G$ with the adjacency matrix $A$. I know that the number of paths of the length $n$ is the sum of elements $A^n$.
But what if we can't walk through a vertex more than one times?

[This is a counter-example to wece's answer, posted only as an answer because it's too long for a comment.]
wece's method is only correct for $A_2$. Counterexample: Let $A=\left[\begin{array}{ccccc}0&1&0&0&0\\1&0&1&0&0\\0&1&0&1&0\\0&0&1&0&1\\0&0&0&1&0\end{array}\right]$ representing a chain of 5 vertices connected by 4 edges. $A_1=A-diag(A)=A$.
$A_2=A\otimes A_1 =\left[\begin{array}{ccccc}1&0&1&0&0\\0&2&0&1&0\\1&0&2&0&1\\0&1&0&2&0\\0&0&1&0&1\end{array}\right]$.
In $A_2$ we see all the paths of length 2, the diagonal representing forbidden paths of length 2 from each vertex back to itself, the off-diagonal entries representing the 8 paths of length 2 (path from $a \rightarrow b$ distinct from $b \rightarrow a$). Then
$A_3 = A \otimes A_2 = \left[\begin{array}{ccccc}0&1&0&0&0\\1&0&1&0&0\\0&1&0&1&0\\0&0&1&0&1\\0&0&0&1&0\end{array}\right]\left[\begin{array}{ccccc}0&0&1&0&0\\0&0&0&1&0\\1&0&0&0&1\\0&1&0&0&0\\0&0&1&0&0\end{array}\right] = \left[\begin{array}{ccccc}0&0&0&1&0\\1&0&1&0&1\\0&1&0&1&0\\1&0&1&0&1\\0&1&0&0&0\end{array}\right]$,
which according to wece's method represents 10 paths of length 3, when there should be 3 such paths.
