# Possible ways to walk to school

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.

• HINT: Since she always takes the shortest walk of $28$ blocks, she MUST go $12$ blocks east and $16$ blocks north each time. Imagine a string like $EEENENNE....N$ describing one set of the 28 moves she makes. Can you take it from here? Commented Oct 21, 2013 at 17:46
• Practically the same was asked hours ago: math.stackexchange.com/questions/533832/… Commented Oct 21, 2013 at 17:46
• If she meets her dance friend at the friend's place, the number of paths is the number $5$ East and $6$ North, times the number of paths $7$ East and $10$ North. Commented Oct 21, 2013 at 17:52
• @Patrick So there would be 28 choose 12 * 16 choose 16 ways to make her moves?
– Nate
Commented Oct 21, 2013 at 17:54
• To be clear, my hint was for part a). No need to multiply since once you choose the $12$ times she moves east you have implicitly chosen where she moves north. Commented Oct 21, 2013 at 18:01

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.