I am not sure how to approach this problem. Every day, a dance student walks from her home to dance-school, which is located $12$ blocks east and $16$ blocks north from home. She always takes the shortest walk of $28$ blocks.
$a$) How many different walks are possible?
$b$) Suppose that $5$ blocks east and $6$ blocks north of her home lives her dance friend, whom she meets each day on her way to dance school. Now how many walks are possible?
For $a$ I was thinking $P(28,12)\cdot P(28,16)$, but not sure if that would work because you can't walk the furthest block north before the closest block north. And for $b$ I'm not really sure where to start.
Best Answer
For part ($a$) we can view walking one block east as an east-step and walking one block north as a north-step. So, different walks correspond to the different combinations of east-steps and north-steps. Thus the number of ways the student can walk $28$ blocks where she walks $12$ east-steps and $16$ north-steps is ${28\choose 12}={28!\over 12!16!}={28\choose 16}$.
For part ($b$) we will first count the number of ways the girl can walk to her friends house. The number of ways she can walk $11$ blocks to her friends house where she takes $5$ east-steps and $6$ north-steps is ${11\choose 5}={11!\over 5!6!}={11\choose 6}$. Now, she must walk the remaining $17$ blocks where she must take $7$ east-steps and $10$ north-steps. So, she can do this in ${17\choose 7}={17!\over 7!10!}={17\choose 10}$ ways. Thus she can walk to school while stopping at her friends house in ${11!\over 5!6!}\cdot {17!\over 7!10!}$ ways.