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Mathematics-Online problems:

# Solution to the problem of the (previous) week

Problem:

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 , , , and , respectively, when setting off at the intersection Galois Avenue/Gauss Street 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.

Solution:

:
The route from to 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

possibilities.

:
Possible routes from to are
 Markov Ave/Riemann St Markov Ave/Euler St Dirac Ave/Riemann St Dirac Ave/Euler St
Accordingly, we obtain

possibilities.

:
If a driver passes the closed intersections at Cea Place to the left of Cea Place, he has

possibilities.

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

possibilities.

Altogether, these add up to 245 different routes.

:
Similar to case , we here obtain

possibilities.

[problem of the week]