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27 Jun 2008, 02:30
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In certain city, a car is parking at A destination, it will go to B destination along routes that are confined to square grid of x streets and y avenues. How many routes from A to B can the car take that have the minimum possible length?
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27 Jun 2008, 02:44
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Minimum length would need to go always right or up. x moves to do "up" and y moves to do "right" i.e. x+y moves total
Among theses x+y moves, we can choose which of these will be the y "right moves", so number of routes that have the minimum length is:
\(C_{x+y}^y=\frac{(x+y)!}{x!y!}\)
Among theses x+y moves, we can choose which of these will be the y "right moves", so number of routes that have the minimum length is:
\(C_{x+y}^y=\frac{(x+y)!}{x!y!}\)
27 Jun 2008, 08:05
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Unless Oski is using a notation I'm unfamiliar with, I think he has x+y and y upside down on the left side of the equation above. In any case, it's not quite the correct answer.
No matter how you travel from A to B, if you can only go North or East, you must go North x-1 times (not x times), and you must go East y-1 times (and not y times). The journey will take (x-1) + (y-1) = x+y-2 steps in total.
Once we have these numbers, Oski's explanation of how to find the total number of paths is entirely correct. I tend to think in terms of words: any 'word' we can make with precisely x-1 Ns and y-1 Es will define a path from A to B. For example, if we have 4 streets and 3 avenues (and thus need to go North 3 times and East 2 times), the word: NNNEE defines the path 'go North three times, then East twice'. The word EENNN is a different path: 'go East twice, then North three times'. Each word you can make with exactly two Es and exactly 3 Ns defines a path from A to B. So really, the question is asking: how many 'words' can you make using exactly x-1 Ns and exactly y-1 Es. We have x+y-2 letters to fill in, and can choose the slots for the Es in "(x+y-2) choose (y-1)" ways. Thus the answer should be:
(x+y-2)!/[(x-1)!*(y-1)!]
No matter how you travel from A to B, if you can only go North or East, you must go North x-1 times (not x times), and you must go East y-1 times (and not y times). The journey will take (x-1) + (y-1) = x+y-2 steps in total.
Once we have these numbers, Oski's explanation of how to find the total number of paths is entirely correct. I tend to think in terms of words: any 'word' we can make with precisely x-1 Ns and y-1 Es will define a path from A to B. For example, if we have 4 streets and 3 avenues (and thus need to go North 3 times and East 2 times), the word: NNNEE defines the path 'go North three times, then East twice'. The word EENNN is a different path: 'go East twice, then North three times'. Each word you can make with exactly two Es and exactly 3 Ns defines a path from A to B. So really, the question is asking: how many 'words' can you make using exactly x-1 Ns and exactly y-1 Es. We have x+y-2 letters to fill in, and can choose the slots for the Es in "(x+y-2) choose (y-1)" ways. Thus the answer should be:
(x+y-2)!/[(x-1)!*(y-1)!]
