Willy Walks to School (Part II)

Let $O=(0,0)$ , $A=(6,6)$ , $B=(9,10)$ , and $S =(10,12)$ . Willy always cuts across the tennis field, which means that he must first arrive at $A$ from $O$ . His path to $A$ must be some combination of $6$ moves up and $6$ moves to the right. The number of these combinations is given by $\dbinom{12}{6}=924$ . There is only $1$ way he can cut across the field, meaning he must end up at $B$ . From $B$ , Willy must make some combination of $2$ moves up and $1$ move to the right in order to get to $S$ . The number of these combinations is $\dbinom{3}{2}=3$ . The total number of ways for Willy to get from his house to $S$ is therefore $n=924*1*3=2772$ . And finally, $\frac{4}{3}n=\fbox{3696}$ .