The problem involves finding the number of shortest paths from the Chen family's home, located at $(0, 5)$, to the hiking trail at $(5, 0)$, using only rightward and downward movements on a grid.

Step 1: Analyze the movement.

To move from $(0, 5)$ to $(5, 0)$, the family must move 5 units to the right (eastward) and 5 units downward (southward), as these are the changes in the x and y coordinates, respectively. Therefore, any shortest path must consist of exactly 5 rightward moves (R) and 5 downward moves (D).

Step 2: Total number of moves.

In total, the family must make 5 rightward moves and 5 downward moves. These movements can occur in any order, so the task reduces to determining how many distinct ways we can arrange 5 R's and 5 D's in a sequence of 10 moves.

Step 3: Calculate the number of distinct paths.

This is a combinatorial problem where we need to choose 5 positions for the rightward moves (R's) from a total of 10 moves. The number of ways to do this is given by the binomial coefficient:

$\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!}$

Step 4: Simplify the factorials.

Now, simplify the expression:

$10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5!$ $\frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = \frac{30240}{120} = 252$

Step 5: Conclusion.

Thus, the number of shortest paths the Chen family can take is:

$\boxed{252}$

Would you like further details, or have any questions?

Here are 5 related questions:

1. What if the grid were larger, say 6x6, how many shortest paths would there be?
2. How does the formula for combinations apply in real-world routing problems?
3. Can we apply this logic to 3D grids (moving in three directions)?
4. What happens if some of the paths are blocked?
5. How can we generalize this problem to paths with more than two types of moves?

Tip: In combinatorics, whenever you're counting distinct arrangements of moves or objects, the binomial coefficient is a handy tool to calculate possibilities.