Understanding the Grid and Pathfinding

The problem asks us to find the number of shortest paths from a starting point P to an ending point Q. We are given a figure made of nine squares, each with a side length of 1 cm. This forms a larger square grid. The student can only walk along the boundaries of these squares.

Let's visualize the grid. Since there are nine squares arranged in a 3x3 pattern, the total dimensions of the figure are 3 cm by 3 cm. Point P is at the top-left corner, and point Q is at the bottom-right corner of this 3 cm x 3 cm square.

Defining Shortest Paths

For a path to be considered the shortest path on a grid like this, the movement must always progress towards the destination. In this case, from P (top-left) to Q (bottom-right), the student can only move right or down. Moving left or up would increase the path length, making it not a shortest path.

Calculating Required Movements

To get from P to Q, the student needs to cover a total horizontal distance and a total vertical distance:

• Horizontal distance: The grid is 3 cm wide (made of 3 squares side-by-side). So, the student must make 3 moves to the right.
• Vertical distance: The grid is 3 cm tall (made of 3 squares stacked vertically). So, the student must make 3 moves down.

In total, any shortest path will consist of exactly 3 right moves and 3 down moves, making a total of $3 + 3 = 6$ moves.

Applying Combinatorics to Count Paths

Each unique sequence of these 6 moves (3 right, 3 down) represents a distinct shortest path. This is a problem of counting combinations, specifically, finding the number of ways to arrange these moves.

We have a total of 6 positions in the path sequence. We need to choose which of these 6 positions will be the 'right' moves (or alternatively, the 'down' moves).

The number of ways to choose $k$ items from a set of $n$ items is given by the binomial coefficient formula:

$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $

In our case:

• Total number of moves, $n = 6$ (3 right + 3 down).
• Number of 'right' moves (or 'down' moves) to choose, $k = 3$.

So, the total number of shortest paths is:

$ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} $

Step-by-Step Calculation

Let's calculate the value:

$ \frac{6!}{3!3!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times (3 \times 2 \times 1)} $

$ = \frac{720}{6 \times 6} $

$ = \frac{720}{36} $

$ = 20 $

Conclusion

Therefore, there are 20 distinct shortest paths from point P to point Q, moving only along the boundaries of the squares and always progressing towards Q.