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Mathematics-Online problems:

Solution to the problem of the (previous) week


The figure shows the concept of a grid plan for a city, where mathematics is to play an important part.


Determine the number of possibilities to reach each of the intersections marked $ A$, $ B$, $ C$, and $ D$, respectively, when setting off at the intersection Galois Avenue/Gauss Street $ K$ and choosing the shortest possible route. (Routes with the same number of intersections are considered to be of equal length.) Take into account that the 2 shaded intersections of Cea Place are closed.


From K to A there exist possibilities.
From K to B there exist possibilities.
From K to C there exist possibilities.
From K to D there exist possibilities.


$ K \rightarrow A$:
The route from $ K$ to $ A$ is composed of 4 vertical and 3 horizontal sections of streets. The different routes correspond to the number of possibilities we have when choosing 3 horizontal sections out of a total of 7. Hence, we obtain

$\displaystyle \binom{7}{3} = \frac{7\cdot6\cdot5}{2\cdot3} = 35\,$possibilities.$\displaystyle $

$ K \rightarrow B$:
Possible routes from $ K$ to $ B$ are
$ K$ $ \longrightarrow$ $ A$ $ \longrightarrow$ $ B$    
$ K$ $ \longrightarrow$ Markov Ave/Riemann St $ \longrightarrow$ Markov Ave/Euler St $ \longrightarrow$ $ B$
$ K$ $ \longrightarrow$ Dirac Ave/Riemann St $ \longrightarrow$ Dirac Ave/Euler St $ \longrightarrow$ $ B$
Accordingly, we obtain

$\displaystyle 35 + \binom{6}{2} \cdot \binom{6}{1} + \binom{5}{1} \cdot \binom{7}{2} = 35 +
90 + 105 = 230\,$possibilities.$\displaystyle $

$ K \rightarrow C$:
If a driver passes the closed intersections at Cea Place to the left of Cea Place, he has

$\displaystyle \binom{5}{1} \cdot \binom{8}{2} = 140\,$possibilities.$\displaystyle $

If he passes them to the right of Cea Place, there exist

$\displaystyle \binom{6}{2} \cdot \binom{7}{1} = 105\,$possibilities.$\displaystyle $

Altogether, these add up to 245 different routes.

$ K \rightarrow D$:
Similar to case $ K \rightarrow B$, we here obtain

$\displaystyle \binom{6}{2} \cdot \binom{4}{1} + \binom{7}{3} = 60 + 35 =
95\,$possibilities.$\displaystyle $

[problem of the week]