Answer:
The answer is "728 routes"
Step-by-step explanation:
The number of ways to proceed from (0, 0) to (5, 7) is as follows:
[tex]\Rightarrow \frac{(7+5)!} {(7! \times 5!)} \\\\ \Rightarrow \frac{12!}{(7! \times 5! )}\\\\\Rightarrow \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{7! \times 5!}\\\\\Rightarrow \frac{12 \times 11 \times 10 \times 9 \times 8}{5!}\\\\\Rightarrow \frac{12 \times 11 \times 10 \times 9 \times 8}{5\times 4 \times 3 \times 2 \times 1}\\\\\Rightarrow \frac{ 11 \times 9 \times 8}{1}\\\\\Rightarrow 11 \times 9 \times 8\\\\\Rightarrow 11 \times 72\\\\\Rightarrow 792[/tex]
We would like to delete the ways of (0, 0) to (2, 3) That's the resolution:
[tex]\Rightarrow \frac{(3+2)!}{(3!\times2!)} \\\\ \Rightarrow \frac{5!}{(3!\times2!)} \\\\\Rightarrow \frac{5\times 4\times 3! }{(3!\times2 \times 1)} \\\\\Rightarrow \frac{5\times 4}{(2 \times 1)} \\\\\Rightarrow \frac{5\times 2}{1} \\\\\Rightarrow 10 \ ways[/tex]
[tex]So, \\\\ \Rightarrow 792 -10 \\\\ \Rightarrow 782 \ routes[/tex]