Avoiding the accident sites

The streets in a particular city form a rectangular grid. The North-South streets are numbered 1st Street, 2nd Street, etc. from west to east. The West-East streets are numbered 1st Avenue, 2nd Avenue, etc. from south to north.

Martin lives at the intersection of 4th Street and 17th Avenue, and wants to jog to the intersection of 11th Street and 9th Avenue. He does not want to spend more time than necessary, so he will only run east and south.

Also, Martin wants to avoid the accident site at the crossing of 10th Street and 15rd Avenue, as well as that at 7th Street and 12th Avenue.

How many jogging routes are there for Martin to choose from?


The answer is 4279.

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1 solution

WIthout the accident sites, Martin's trip consists of 15 units, of which 7 are east and 8 south. That gives a total of ( 15 7 ) = 6435 \left(\begin{array}{c} 15 \\ 7\end{array}\right) = 6435 routes. We subtract the routes that include an accident site. Fortunately no valid will include both accident sites (why not?), so we can subtract them separately.

Any route to the first accident site consists of 8 units, 6 of which are east and 2 south. To complete the jog, Martin would run another 7 units, of which 1 goes east and 6 south. This accounts for ( 8 6 ) ( 7 1 ) = 196 routes . \left(\begin{array}{c} 8 \\ 6\end{array}\right)\cdot\left(\begin{array}{c} 7 \\ 1\end{array}\right) = 196\ \text{routes}. Likewise, for the second accident site we get ( 8 3 ) ( 7 4 ) = 1960 routes . \left(\begin{array}{c} 8 \\ 3\end{array}\right)\cdot\left(\begin{array}{c} 7 \\ 4\end{array}\right) = 1960\ \text{routes}. That leaves 6435 196 1960 = 4279 6435 - 196 - 1960 = \boxed{4279} routes that avoid the accident sites.

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