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The following figure depicts the paths from home to work. SAM never travels through the park when going to work.

enter image description here

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manual counting gives 8 ways. Is there a mathematical process which can give the answer without manual counting. – Rajesh K Singh Feb 5 '13 at 16:11
Counting is a mathematical process. – Henning Makholm Feb 5 '13 at 16:28
up vote 2 down vote accepted

He has to pass through exactly one of E, F, and C. If he passes through E or C then there's trivially only one path (for each) he can take. If he passes through F, there are $\binom{2}{1} = 2$ ways he can get from A to F and $\binom{3}{2} = 3$ ways he can get from F to K, for a total of $2\times 3 = 6$. So in all there are $1+1+6 = 8$ ways he can get home.

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Write the number of possibilities of reaching a vertex $X$ from $A$. It starts like $A:1$ as getting to $A$ is unique, both $B$ and $D$ also have only $1$ possibilities. They simply add up, for example the possibilities to arrive at $I$ are the possibilities arriving at either to $H$ or to $G$ (plus the unique $GI$ path, resp. $HI$ path). $$A:1,\ B:1,\ D:1,\ F:2,\ E:1,\ G:3,\ H:2,\ I:5,\ C:1,\ J:3,\ K:8 $$

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If the park were not there, there would be ${3+2 \choose 2}=10$ possible paths.

Adding the park cuts off one possible line segment Sam could walk along (from the $BC$ midpoint to $H$). There is only one way he could get to the top of that line segment, and $2$ possible continuances once he reaches $H$. So the total number of paths that don't go through the park is $10 - 1 \cdot 2 = 8$.

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