Willy Avoids Fences

The problem statement itself has one major error I guess.

He needs to get to school by making successive movements one unit to the left or one unit up.

It should be one unit to the right or one unit up , because otherwise,it's impossible to go to the topmost right corner of the grid from the origin.

And the rest is simple. First count the number of ways from (0,0) to (6,4) . It should be 10C4 = 210. Then,count the number of ways from (0,0) to (1,1) and (2,1) to (6,4) . It should be 2C1 7C3 = 70. Then,count the number of ways from (0,0) to (3,4) and (4,4) to (6,4) . It should be 7C3 2C0 = 35. Then,count the number of ways from (0,0) to (1,1) and (2,1) to (3,4) and (4,4) to (6,4) . It should be 2C1 4C1 2C0 = 8,which should be subtracted.

Hence,the final count is = 210-(70+35-8) = 113

There are $\binom{10}{6} = 210$ ways for Willy to get from $(0,\,0)$ to $(6,\,4)$ that go through either wall (possibly both , possibly neither ).

There are $\binom{2}{1} \cdot \binom{7}{3} = 2 \cdot 35$ ways for Willy to get from $(0,\,0)$ to $(6,\,4)$ that go through the first wall.

There are $\binom{7}{3} \cdot \binom{2}{0} = 35 \cdot 1$ ways for Willy to get from $(0,\,0)$ to $(6,\,4)$ that go through the second wall.

There are $\binom{2}{1} \cdot \binom{4}{1} \cdot \binom{2}{0} = 2 \cdot 4 \cdot 1$ ways for Willy to get from $(0,\,0)$ to $(6,\,4)$ that go through both walls.

Therefore, there are $210 - 2 \cdot 35 - 35 \cdot 1 + 2 \cdot 4 \cdot 1 = 210 - 70 - 35 + 8 = \boxed{113}$ ways for Willy to get from $(0,\,0)$ to $(6,\,4)$ that go through neither walls.