To solve this problem, we need to determine the number of different paths from point $A$ to point $B$, given that each path involves moving exactly 4 steps to the right and 3 steps up.

Solution Approach

Since we are restricted to moving only up or to the right, each path from $A$ to $B$ will consist of exactly:

• 4 steps to the right
• 3 steps upward

Thus, each valid path is a sequence of 7 steps (4 "Right" and 3 "Up").

Counting the Paths

The task is essentially to find the number of distinct arrangements of 4 "Right" moves and 3 "Up" moves in a sequence of 7 moves. This is a combinatorial problem, where we are choosing 4 positions out of 7 for the "Right" moves (or equivalently, choosing 3 positions out of 7 for the "Up" moves).

The number of distinct arrangements can be calculated using the binomial coefficient:

$\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4! \cdot 3!}$

Calculating this:

1. $7! = 5040$
2. $4! = 24$
3. $3! = 6$

So,

$\binom{7}{4} = \frac{5040}{24 \cdot 6} = \frac{5040}{144} = 35$

Answer

There are 35 different paths from point $A$ to point $B$.

Would you like a breakdown of these steps or an explanation on how to calculate binomial coefficients? Here are related questions to explore:

1. How would the answer change if the grid size increased to 5 steps to the right and 3 steps up?
2. What is the general formula for counting paths in an $m \times n$ grid?
3. How would the problem change if diagonal moves were allowed?
4. Could we use Pascal’s Triangle to find the solution?
5. What is the relationship between this problem and combinatorial probability?

Tip: In combinatorics, breaking down complex movements into sequences of binary choices (like "Right" or "Up") often simplifies the path-counting problem.