Combinations for a grid I was wondering if I could get some help with this question. I've made an attempt at solving for the total number of paths but I can't quite grasp how to account for the blocked out squares. If anyone could explain this it would be much appreciated. Thank you!
"Determine the number of possible routes from Adam's house to Matt's house if you can only travel south and/or east. Hint: you can only travel when there is a line connecting."

This is what I've done so far:
There are 6 blocks east and 5 blocks south. 6+5=11.
11!/(6!5!)=462.
I think this is the total number of paths, but I don't know how to account for the blocked squares.
 A: Notice that you cannot cross in two of the blocked squares on the same walk. Just one! So just do $$\text{Total number of paths}-\left (\sum _{i=1}^3\text{paths going thru i-th blocked}\right ).$$
The total sum you have already done. For going thru the $i-$th block, just go from Adam's to the start of the block and multiply it(why?) to the number of ways to go from the end of the block to Matt's.

$$\binom{6+5}{5}-\binom{4+1}{1}\binom{2+3}{2}-\binom{2+3}{2}\binom{4+1}{1}-\binom{3+3}{3}\binom{3+1}{1}$$

A: To my mind the easiest way to do problems like this is to write down more generally, for each grid point, the number of ways to get to that grid point. The advantage of doing this is that they satisfy a very simple recurrence generalizing Pascal's rule for Pascal's triangle: the number on a point is the sum of the numbers to its north and west, except you ignore a number if the relevant path is blocked. This is what the calculation looks like and we get $282$ as you calculated:

Doing the calculation this way is not much more difficult even with a pretty big grid and a complicated pattern of blocked paths. Also it's quite easy to check your work since you just need to check every application of the modified Pascal's rule.
