To solve this, we need to determine how many paths exist from $(2, 3)$ to $(5, 4)$ and then from $(5, 4)$ to $(9, 7)$ on a grid, given that movement is restricted to either going right $(x + 1, y)$ or going up $(x, y + 1)$.

Step 1: Paths from $(2, 3)$ to $(5, 4)$

• From $(2, 3)$ to $(5, 4)$, the total change in $x$ is $5 - 2 = 3$ steps right.
• The total change in $y$ is $4 - 3 = 1$ step up.
• The number of paths is the number of ways to arrange 3 rights (R) and 1 up (U) in any order. This is given by the combination formula: $\text{Number of paths} = \binom{3+1}{1} = \binom{4}{1} = 4$

Step 2: Paths from $(5, 4)$ to $(9, 7)$

• From $(5, 4)$ to $(9, 7)$, the total change in $x$ is $9 - 5 = 4$ steps right.
• The total change in $y$ is $7 - 4 = 3$ steps up.
• The number of paths is the number of ways to arrange 4 rights (R) and 3 ups (U), which is: $\text{Number of paths} = \binom{4+3}{3} = \binom{7}{3} = 35$

Step 3: Total Paths

The total number of paths from $(2, 3)$ to $(9, 7)$, passing through $(5, 4)$, is the product of the two values: $\text{Total paths} = 4 \times 35 = 140$

Thus, the answer is: $\boxed{140}$

Would you like any further details or clarifications on this problem?

5 Related Questions:

1. How would the number of paths change if you had to pass through multiple intermediate points?
2. How do binomial coefficients help in counting grid paths?
3. Can this method be extended to higher dimensions?
4. What is the general formula for finding the number of paths between two points in a grid?
5. How would obstacles in the grid affect the number of possible paths?

Tip: The combination formula $\binom{n}{k}$ helps in counting arrangements where order matters but repetition is not allowed, which is essential in path-counting problems.