Number of paths across a grid (extended)

Question

http://furthermaths.org.uk/docs/Groupfinal1718.pdf

Question 4 part (a) is the classic shortest route across a grid question. Part (b) involves a twist; how is it done?

Answer 1

Notice that the shortest route from $A$ to $B$ takes five steps, of which exactly three are to the right and two are upwards.

An admissible path of length $7$ can occur in two ways:

1. It involves exactly four steps to the right, one to the left, and two upward steps.
2. It involves exactly three upward steps, one downward step, and three steps to the right.

The case in which there are exactly four steps to the right, one to the left, and two upward steps can occur in $$\binom{7}{4, 1, 2} = \frac{7!}{4!1!2!}$$ ways. The case in which there are exactly three upward steps, one downward step, and three steps to the right can occur in $$\binom{7}{3, 1, 3} = \frac{7!}{3!1!3!}$$ ways. Since these cases are mutually exclusive and exhaustive, the number of admissible paths is $$\binom{7}{4, 1, 2} + \binom{7}{3, 1, 3}$$