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Practical applications of combinatorics
The following examples raise a few problems that arise while skiing but need mathematical considerations. Besides solving the examples it is also worth thinking about further questions that can be raised.
Example 1
The ski-lifts of a ski-run are denoted by blue arrows and the slopes are denoted by red arrows in Figure 57. The table contains the data of the ski-lifts.
- How can we get from the point \latex{B} to the point \latex{F}, if we either go upwards with a ski-lift or we slide downwards on skis?
- What can be the difference of heights between the points \latex{B} and \latex{F}, and the points \latex{A} and \latex{G}?
- In how many different ways can we get down from \latex{G} to the point \latex{A}, if we only slide downwards?
- What is the angle of elevation on the ski-lift \latex{AB}?
Figure 57
\latex{ C }
\latex{ D }
\latex{ J }
\latex{ I }
\latex{ F }
\latex{ H }
\latex{ G }
\latex{ E }
\latex{ B }
\latex{ A }
Let us examine where the point \latex{C} can be so that we can slide down there both from \latex{B} and from \latex{E}. The point \latex{B} is \latex{547\,m} higher and the point \latex{E} is \latex{380 + 258 = 638\,m} higher, thus the point \latex{C} is slightly lower than these, i.e. it is higher than \latex{A} by at most \latex{547\,m}.
Solution (a)
Based on the arrows it can be seen that we can get from \latex{B} to \latex{F} if we slide down to \latex{C}, then we take the ski-lift to \latex{G}, then we slide down to \latex{F} through \latex{H}. In the meantime we can ski the cycle \latex{BCAB}. We can go from \latex{H} also to \latex{C}, from here up to \latex{G} or down to \latex{A}, thus by adding the cycle \latex{CABC} or the cycle \latex{CADEC}.
length
difference
of heights
of heights
\latex{ AB }
\latex{ AD }
\latex{ DE }
\latex{ CG }
\latex{ FE }
\latex{1,105\,m}
\latex{1,200\,m}
\latex{820\,m}
\latex{420\,m}
\latex{547\,m}
\latex{867\,m}
\latex{380\,m}
\latex{258\,m}
\latex{149\,m}
\latex{258\,m}
Solution (b)
The differences of heights between the points can be given based on the ski-lifts: The point \latex{B} is higher than \latex{A} by \latex{547\,m}, the point \latex{E} is higher by \latex{638\,m}, the point \latex{F} is lower than it by \latex{258\,m}, thus the point \latex{F} is higher than \latex{A} by \latex{638 – 258 = 380\,m}. So the point \latex{B} is higher than \latex{F} by \latex{547 – 380 = 167\,m}.
We only know about the point \latex{C} that it is higher than \latex{A}, but it cannot be higher than \latex{A} by \latex{547\,m}. Since \latex{G} is higher than \latex{C} by \latex{149\,m}, thus \latex{G} is higher than \latex{A} by at most \latex{547 + 149 = 696\,m}. \latex{G} is higher than \latex{E}, which is \latex{638\,m} higher above \latex{A}, thus the difference of heights between \latex{A} and \latex{G} is at least \latex{638\,m}, but it is at most \latex{696\,m}. (Figure 58)
Figure 58
difference of heights
\latex{ 500 }
\latex{ 547 }
\latex{ 638 }
\latex{ 258 }
\latex{ 149 }
A
B
C
E
F
G
Solution (c)
Let us highlight in green the edges of the graph, which lead downwards from \latex{G} towards \latex{A}, and based on the summation method let us write next to the points the number of the different ways we can get to the point starting from \latex{G}. (Figure 59)
We can get to \latex{H} in \latex{1} way, just like further to \latex{E}. There is \latex{1} path leading to \latex{I}, \latex{2} paths to \latex{J}, thus there are \latex{3} paths leading to \latex{D}. We can get to \latex{C} from \latex{H} or from \latex{E}, thus \latex{2} paths lead here, finally we can slide down to \latex{A} from \latex{C} or from \latex{D}, thus we can get down to \latex{A} from \latex{G} in \latex{5} different ways.
Figure 59
\latex{ E }(\latex{ 1 })
\latex{ C }(\latex{ 2 })
\latex{ H }(\latex{ 1 })
\latex{ I }(\latex{ 1 })
\latex{ J }(\latex{ 2 })
\latex{ D }(\latex{ 3 })
\latex{ A }
\latex{ F }
\latex{ G }
\latex{ B }
Solution (d)
The elevation of the ski-lift \latex{AB} is \latex{547\,m}, its length is \latex{1,105\,m}, thus the following can be written for the angle of elevation:
\latex{\sin \alpha=\frac {547}{1,105}=0.4950}, which implies \latex{\alpha=29.67\degree}.
So the angle of elevation on the ski-lift \latex{AB} is almost \latex{30\degree}. Let us ask more questions and let us answer them.
Exercises
{{exercise_number}}. The table on the left shows the data of the tourist hostel \latex{600\,km} away from Budapest, next to the ski resort (prices for \latex{6} nights and \latex{1} person in EUR), and the table on the right shows the insurance prices:
\latex{ 30/9–23/12 }
\latex{ 24/12–5/1 }
\latex{ 6/1–8/3 }
\latex{ 9/3–2/4 }
room for \latex{ 2 }
room for \latex{ 3 }
room for \latex{ 4 }
adult
child
family
\latex{(2+2)}
\latex{(2+2)}
\latex{ 1.25 } EUR/day
\latex{ 0.85 } EUR/day
\latex{ 3.40 } EUR/day
\latex{ 73 }
\latex{ 67 }
\latex{ 61 }
\latex{ 106 }
\latex{ 73 }
\latex{ 90 }
\latex{ 100 }
\latex{ 83 }
\latex{ 67 }
\latex{ 94 }
\latex{ 80 }
\latex{ 61 }
The price of half board for \latex{1} person is \latex{7} euros daily.
It is free for children under \latex{2} years, \latex{50\% } of the price of the bed and meal should be paid for \latex{2}–\latex{6}-year-old children, and the \latex{75\%} of the price should be paid for \latex{6}-\latex{14}-year-old children.
Parking a car for a day in the car park of the tourist hostel costs \latex{5} euros.
A family of four from Budapest (\latex{2} adults, a \latex{10}-year-old and a \latex{5}-year-old child) go skiing by car in February for a week (\latex{6} nights). They arrive at noon on the first day, then they ski, have dinner, and they leave for home after breakfast on the last day. How much does the skiing costs them without the ski passes, but with half board, if the average fuel consumption of their car is \latex{6.8} litre (\latex{1.35} EUR/litre) on \latex{100\,km}?
Parking a car for a day in the car park of the tourist hostel costs \latex{5} euros.
A family of four from Budapest (\latex{2} adults, a \latex{10}-year-old and a \latex{5}-year-old child) go skiing by car in February for a week (\latex{6} nights). They arrive at noon on the first day, then they ski, have dinner, and they leave for home after breakfast on the last day. How much does the skiing costs them without the ski passes, but with half board, if the average fuel consumption of their car is \latex{6.8} litre (\latex{1.35} EUR/litre) on \latex{100\,km}?
Prices of the ski pass in euros:
adult
junior (\latex{ 14–18 } years)
child (\latex{ 6–14 } years)
senior over (\latex{ 65 } years)
daily
pass
pass
after
\latex{ 1 } pm
\latex{ 1 } pm
\latex{ 2 } days
\latex{ 3 } days
\latex{ 4 } days
\latex{ 5 } days
\latex{ 6 } days
\latex{ 7 } days
\latex{ 26.5 }
\latex{ 23.9 }
\latex{ 15.1 }
\latex{ 24.4 }
\latex{ 21.7 }
\latex{ 50.9 }
\latex{ 72.4 }
\latex{ 84.7 }
\latex{ 101.9 }
\latex{ 117.2 }
\latex{ 131.3 }
\latex{ 19.6 }
\latex{ 45.8 }
\latex{ 65.2 }
\latex{ 76.3 }
\latex{ 91.7 }
\latex{ 105.5 }
\latex{ 118.1 }
\latex{ 12.4 }
\latex{ 29.0 }
\latex{ 41.3 }
\latex{ 48.3 }
\latex{ 58.1 }
\latex{ 66.8 }
\latex{ 74.8 }
\latex{ 20.0 }
\latex{ 46.8 }
\latex{ 66.6 }
\latex{ 78.0 }
\latex{ 93.8 }
\latex{ 107.8 }
\latex{ 120.8 }
At least how much do the ski passes cost for the family if all of them ski all week long? How much risk do they take if they buy the passes in advance, however according to the weather forecast there will be melting on the last two days?