There are five cities A, B, C, D, E on a certain island. Each city is connected to every other city by road. In how many ways can a person starting from city A come back to A after visiting some cities without visiting a city more than once and without taking the same road more than once ? (The order in which he visits the cities also matters. e.g., the routes `A rarr B rarr C A and A rarr C rarr B rarr A` are different).

Answer

To solve the problem of how many ways a person can start from city A, visit some cities, and return to A without visiting any city more than once and without taking the same road more than once, we can break down the solution into several cases based on the number of cities visited.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have 5 cities: A, B, C, D, and E.
- The person starts at city A and can visit any combination of the other cities (B, C, D, E) before returning to A.
- The order of visiting matters, and no city can be visited more than once.

2. **Case 1: Visiting 4 Cities**:
- If the person visits 4 cities (B, C, D, E), they will visit all of them before returning to A.
- The number of ways to arrange 4 cities is given by \(4!\) (4 factorial).
- Calculation:
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]

3. **Case 2: Visiting 3 Cities**:
- The person can choose any 3 cities from the 4 available (B, C, D, E).
- The number of ways to choose 3 cities from 4 is given by \(\binom{4}{3}\), and the number of arrangements of these 3 cities is \(3!\).
- Calculation:
\[ \binom{4}{3} = 4 \quad \text{(ways to choose 3 cities)} \]
\[ 3! = 6 \quad \text{(ways to arrange 3 cities)} \]
\[ \text{Total for this case} = 4 \times 6 = 24 \]

4. **Case 3: Visiting 2 Cities**:
- The person can choose any 2 cities from the 4 available (B, C, D, E).
- The number of ways to choose 2 cities from 4 is given by \(\binom{4}{2}\), and the number of arrangements of these 2 cities is \(2!\).
- Calculation:
\[ \binom{4}{2} = 6 \quad \text{(ways to choose 2 cities)} \]
\[ 2! = 2 \quad \text{(ways to arrange 2 cities)} \]
\[ \text{Total for this case} = 6 \times 2 = 12 \]

5. **Total Number of Ways**:
- Now, we sum the total ways from all cases:
\[ \text{Total} = 24 \, (\text{from case 1}) + 24 \, (\text{from case 2}) + 12 \, (\text{from case 3}) = 60 \]

### Final Answer:
The total number of ways a person can start from city A, visit some cities, and return to A is **60**.