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Mathematics-Online problems: | ||

## Solution to the problem of the (previous) week |

** Problem:**

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.

**Answer:**

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 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 existpossibilities.Altogether, these add up to 245 different routes.

- :
- Similar to case
, we here obtain
possibilities.

[problem of the week]