To solve this problem, we need to find the number of shortest paths from point $A$ to point $R$ in the grid. Here is how we can approach it:

Observing the Grid:

• We can only move down or right along the lines in the grid.
• The shortest path consists of the minimum number of moves, where each move takes us closer to $R$ either horizontally or vertically.
• We need to calculate how many down and right moves are required to get from $A$ to $R$.

Step 1: Number of Moves

• From $A$, we need to move down and right to get to $R$.

• Count the vertical lines from $A$ to $R$:

  • There are 3 vertical moves (3 "down" steps).

• Count the horizontal lines from $A$ to $R$:

  • There are 4 horizontal moves (4 "right" steps).

Thus, the shortest path requires exactly 3 down moves and 4 right moves.

Step 2: Total Number of Paths

The number of different paths can be found by determining how many ways we can arrange these moves. Since we are making a total of $7$ moves (3 downs and 4 rights), the problem boils down to choosing 3 positions for the "down" moves out of 7.

This is a combinatorial problem, and the number of such arrangements is given by the combination formula: $\text{Number of paths} = \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

Final Answer:

There are 35 different shortest paths from point $A$ to point $R$.

Would you like more details on the combinatorics involved or have any other questions?

Here are 5 questions related to this concept:

1. What is the general formula for the number of shortest paths on a grid?
2. How does the number of paths change if we increase the grid size?
3. Can this method be applied to three-dimensional grids? If so, how?
4. What if diagonal moves were allowed in the grid—how would that affect the calculation?
5. How does this concept connect to Pascal's triangle?

Tip: In combinatorics problems, identifying the number of each type of required move (right/down) and their arrangement is key to calculating the number of possible paths.