The Chen family wants to go for a hike. Before they leave, they realize they don't have any snacks to bring with them, so they want to stop at the market to grab some on the way to the trail. In how many ways can they drive through the city from their home to the trail, if they want to take the shortest route possible and stop by the market? (If there is more than one such route, the family is OK with taking any such route.)

[asy]
unitsize(1cm);
add(grid(5,5));
label("Home",(0,5),NW);
dot((0,5));
label("Trail",(5,0),SE);
dot((5,0));
label(scale(0.7)*rotate(45)*"Market",(2,3),NE);
dot((2,3));
draw("$N$",(6,1.0)--(6,2.0),5N,EndArrow(4));
draw((5.8,1.3)--(6.2,1.3));
[/asy]

The problem can be modeled on a grid where the family needs to move from their home to the trail, making sure to stop by the market on the way. The movement on the grid is restricted to the shortest path, meaning they can only move right (east) or down (south).

### Step 1: Grid Coordinates

The home is at $ (0, 5) $, the trail is at $ (5, 0) $, and the market is at $ (2, 3) $.

### Step 2: Path from Home to Market

To go from home $ (0, 5) $ to the market $ (2, 3) $, the family needs to move:
- 2 steps to the right (east),
- 2 steps down (south).

This is equivalent to finding how many ways we can arrange 2 "east" (E) steps and 2 "south" (S) steps. The total number of steps is 4, and the number of ways to choose 2 east steps out of 4 is given by the binomial coefficient:
\[
\binom{4}{2} = \frac{4!}{2!2!} = 6.
\]
Thus, there are 6 ways to go from home to the market.

### Step 3: Path from Market to Trail

To go from the market $ (2, 3) $ to the trail $ (5, 0) $, the family needs to move:
- 3 steps to the right (east),
- 3 steps down (south).

Similarly, the total number of steps is 6, and the number of ways to arrange 3 east steps out of 6 is:
\[
\binom{6}{3} = \frac{6!}{3!3!} = 20.
\]
Thus, there are 20 ways to go from the market to the trail.

### Step 4: Total Number of Ways

To find the total number of ways to drive from home to the trail, stopping at the market, we multiply the number of ways for each leg of the trip:
\[
6 \times 20 = 120.
\]

Thus, the family can drive from their home to the trail, stopping by the market, in **120 different ways**.

---

Do you want further details, or have any questions? Here are 5 related problems to explore:

1. What if the market is located at $ (3, 2) $ instead? How many ways would there be?
2. How many ways are there if they take a detour to another location after the market?
3. If they need to take exactly one extra step to sightsee, how many total paths could they take?
4. What happens if they can also move diagonally along the grid?
5. How would the problem change if they could only move east?

**Tip:** Binomial coefficients are a powerful way to count combinations and arrange steps in problems like this. Understanding how they work can simplify many counting problems.

The Chen family wants to go for a hike. Before they leave, they realize they don't have any snacks to bring with them, so they want to stop at the market to grab some on the way to the trail. In how many ways can they drive through the city from their home to the trail, if they want to take the shortest route possible and stop by the market?

Learn how to calculate the number of ways the Chen family can drive from their home to the trail, stopping by the market using grid path combinatorics. The problem involves using binomial coefficients and combinatorial mathematics, making it suitable for students in Grades 9-12.

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