Is this for Math Challenge II-B Winter session (Combinatorics)?

If so, the question 5.23 is the following:

Suppose an ant starts at the origin $(0,0)$. Ever step it takes is either $(1,1)$ or $(1,−1)$ (so it moves diagonally up and diagonally down).

(a) How many different ways can the ant move from the origin to $(20, 0)$?

This question is similar to question 5.3.  To reach the point $(20,0)$, the ant has to make $10$ steps up and $10$ steps down.  In total there are $20$ steps, and the up and down steps can be in any order.  Thus, we are counting the number of ways to choose $10$ out of $20$, and the answer is $\binom{20}{10}$.

Part (b) repeats part (a) with the requirement that the ant stop at the point $(10,0)$.  This a two-part process: the first part is from $(0,0)$ to $(10,0)$, and the second part is from $(10,0)$ to $(20,0)$.  Each part is similar to Part (a), and the result is the product of the two parts.