17) The figure on the side is the plan of a neighborhood: points $A, B, C$ and $D$ are the houses and the segments are the streets. From how many of the four houses is it possible to start a path that passes exactly once through each street (possibly passing more than once through the same house)?
(A) 0
(B) 1,
(C) 2, (D) 3
(E) 4.

![](https://cdn.mathpix.com/cropped/2024_04_17_fd905e9b1ee39ff4ae71g-2.jpg?height=202&width=197&top_left_y=1253&top_left_x=1274)
17. The answer is $(\mathbf{C})$.

Consider a path that starts from $B$; suppose the first road taken is $B A$. If we then go to $C$, we cannot go to $B$ afterward because we would be stuck without having traversed all the roads. If instead we go to $D$, we are then forced to go to $A$, again getting stuck without having traversed the road $B C$. Similarly (the reasoning is entirely symmetric), it can be shown that if we start from $B$ and go to $C$, it is not possible to complete the path with the required characteristics. By symmetry, this also proves that there are no paths with the required characteristics that start from $D$. The path that starts from $A$:

$$
A \rightarrow C \rightarrow B \rightarrow A \rightarrow D \rightarrow C
$$

has the required characteristics. Similarly, the path that starts from $C$:

$$
C \rightarrow A \rightarrow B \rightarrow C \rightarrow D \rightarrow A
$$

has the required properties. In conclusion, there are paths with the required characteristics only when starting from $A$ and from $C$.