Residents of planet Earthella are planning to abandon it and fly to planet Marsella because of an acute water crisis. They may choose any of the flight paths connecting Earthella to Marsella through a set of intermediate refuelling stations. How many unique flight paths be charted out if paths cannot retrace any segment that they have already traversed or revisit any station (including Earthella)?
Show Hint
For path-counting problems in complex graphs, be very systematic. Breaking the problem into disjoint cases based on the first step is a great strategy. Using a tree diagram to trace paths from the start node can help ensure you don't miss any or count any twice. Always check for symmetry in the graph to save time.
Step 1: Understanding the Question
The problem asks for the total number of unique simple paths from the start station (Earthella) to the end station (Marsella). A simple path is one that does not revisit any station (vertex) or traverse any segment (edge) more than once. We need to systematically count all such paths in the given network diagram. Step 2: Key Formula or Approach
This is a path-finding problem on a graph. The most reliable method for a small graph is a systematic enumeration of all possible paths. We can break down the problem by considering the first station visited after leaving Earthella. Earthella (E) is connected to three stations: a top station (let's call it T1), a central station (C1), and a bottom station (B1).
The total number of paths will be the sum of:
1. Paths starting with the segment E \(\rightarrow\) T1.
2. Paths starting with the segment E \(\rightarrow\) B1.
3. Paths starting with the segment E \(\rightarrow\) C1.
Due to the symmetry of the graph along the horizontal axis, the number of paths starting with E \(\rightarrow\) T1 will be equal to the number of paths starting with E \(\rightarrow\) B1. Step 3: Detailed Explanation
Let's label the stations as E (Earthella), M (Marsella), T1, T2 (top stations), B1, B2 (bottom stations), and C1 (central station). Case 1: Paths starting with E \(\rightarrow\) T1
We list all unique paths from T1 to M without revisiting E or T1.
1. E \(\rightarrow\) T1 \(\rightarrow\) T2 \(\rightarrow\) M
2. E \(\rightarrow\) T1 \(\rightarrow\) T2 \(\rightarrow\) C1 \(\rightarrow\) M
3. E \(\rightarrow\) T1 \(\rightarrow\) T2 \(\rightarrow\) C1 \(\rightarrow\) B2 \(\rightarrow\) M
4. E \(\rightarrow\) T1 \(\rightarrow\) T2 \(\rightarrow\) C1 \(\rightarrow\) B1 \(\rightarrow\) B2 \(\rightarrow\) M
5. E \(\rightarrow\) T1 \(\rightarrow\) C1 \(\rightarrow\) M
6. E \(\rightarrow\) T1 \(\rightarrow\) C1 \(\rightarrow\) T2 \(\rightarrow\) M
7. E \(\rightarrow\) T1 \(\rightarrow\) C1 \(\rightarrow\) B2 \(\rightarrow\) M
8. E \(\rightarrow\) T1 \(\rightarrow\) C1 \(\rightarrow\) B1 \(\rightarrow\) B2 \(\rightarrow\) M
There are 8 paths in this case. Case 2: Paths starting with E \(\rightarrow\) B1
Due to symmetry, the number of paths starting with E \(\rightarrow\) B1 is the same as the number of paths starting with E \(\rightarrow\) T1.
So, there are 8 paths in this case as well. Case 3: Paths starting with E \(\rightarrow\) C1
We list all unique paths from C1 to M without revisiting E or C1. The next station cannot be T1 or B1, as those would be counted as paths starting with E \(\rightarrow\) T1 \(\rightarrow\) C1... or E \(\rightarrow\) B1 \(\rightarrow\) C1... in the previous cases. However, our main grouping is based on the first edge, so we must count all paths starting with E \(\rightarrow\) C1.
1. E \(\rightarrow\) C1 \(\rightarrow\) M
2. E \(\rightarrow\) C1 \(\rightarrow\) T2 \(\rightarrow\) M
3. E \(\rightarrow\) C1 \(\rightarrow\) T1 \(\rightarrow\) T2 \(\rightarrow\) M
4. E \(\rightarrow\) C1 \(\rightarrow\) B2 \(\rightarrow\) M
5. E \(\rightarrow\) C1 \(\rightarrow\) B1 \(\rightarrow\) B2 \(\rightarrow\) M
There are 5 paths in this case. Step 4: Final Answer
The total number of unique paths is the sum of the paths from all three cases, as they are mutually exclusive based on the first segment of the journey.
Total Paths = (Paths from Case 1) + (Paths from Case 2) + (Paths from Case 3)
\[ \text{Total Paths} = 8 + 8 + 5 = 21 \]
There are 21 unique flight paths from Earthella to Marsella.
Therefore, the correct option is (D).
