Street combinatorics problem A secretary works in a building located nine blocks east and eight blocks north of his home.  Every day he walks 17 blocks to work.  The map is a 9 x 8 and each block is a square of the same size.
1) How many different routes are possible for him?  $\binom{17}{9}$
2) How many different routes are possible if the one block in the easterly direction, whic begins four blocks east and three blocks north of his home, is under water (and he can't swim) (Hint: Count the routes that use the block under water)
I solved the first one on my own. Can anyone help me with the second question?  
 A: Well, the assumption is he only travels north and east and never west and south.. this should be made clear.
Hint: Now, if the block is under water, presumably he cannot use any of the streets around the block. Let us say the block is the square with vertices $(m,n)$, $(m+1,n)$, $(m, n+1)$, $(m+1, n+1)$. 
Then, the number of routes that include a street near the block are those that pass through $(m,n)$ plus those that pass through $(m+1, n)$ and dont pass through $(m,n)$ plus those that pass through $(m, n+1)$ and dont pass through $(m,n)$.. 
[Note that any route that passes through $(m+1,n+1)$ must also pass through some other vertex, so you dont need to count these].
Can you solve it now?
A: You have correctly calculated the number of ways the secretary can travel to work from his home by walking $17$ blocks. 
In the diagram below, the flooded block is shown in blue.

We must subtract the number of routes that pass through the flooded block from the total number of ways the secretary can walk to his work by traveling $17$ blocks.
Notice that in all such routes, the secretary must first travel four blocks east and three blocks north, traverse the flooded block, then travel an additional four blocks east and five blocks north.
Can you take it from here?
A: He only walks through the water when walking exactly three blocks north and four
blocks east in the first seven moves then walking east in the eight move.
There are $\frac{7!}{4!3!}$ ways for him to be at the intersection with water to
the east (permutations of a multiset with 3 N's and 4 E's). There is one way for him to travel for him through the water (east)
and $\frac{9!}{4!5!}$ ways for him to travel the remaining distance.
There are $\frac{17!}{8!9!} - \frac{7!}{4!3!} \frac{9!}{4!5!}$ ways for him to
walk home without going through the water.
