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JamesJ

  • 4 years ago

For Avavind: There are K intermediate stations on a railway line from one terminus to another. In how many ways can a train stop at three of these intermediate stations if no two of these stopping stations are to be consecutive?

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  1. JamesJ
    • 4 years ago
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    This is tricky problem. Obviously if we don't have constraints of no consecutives, then there are \[ {k \choose 3} \] choices. But with the constraint, it get's trickier. Let's look at a few examples. If k = 5, and we label the stations 1 2 3 4 5, then there is only one option: 1 3 5 If k = 6, then there are four options: 1 3 5 1 3 6 1 4 6 2 4 6 If k=7, there are ten options: 1 3 5 1 3 6 1 3 7 1 4 6 1 4 7 1 5 7 2 4 6 2 4 7 2 5 7 3 5 7 If k=8, there are twenty options.

  2. JamesJ
    • 4 years ago
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    Now how to generalize ...

  3. anonymous
    • 4 years ago
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    James Please hold on sec, I want to give it a fair try :)

  4. JamesJ
    • 4 years ago
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    Sure.

  5. anonymous
    • 4 years ago
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    Guessing from pattern: \( \large \binom{K-2}{3} \)

  6. anonymous
    • 4 years ago
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    Do you have a solution? @james

  7. JamesJ
    • 4 years ago
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    I do.

  8. JamesJ
    • 4 years ago
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    Intuitively, your hypothesis makes sense.

  9. anonymous
    • 4 years ago
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    Okay, so what's your analysis ? :)

  10. JamesJ
    • 4 years ago
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    Suppose the three stations in order are located at a, b and c. We know that a must be at least 4 stations away from the last station, hence \[ 1 \leq a \leq k - 4\] Now b must be at least 2 stations after a and at least 2 stations before the k-th: \[ a + 2 \leq b \leq k - 2 \] Similarly, c must be at least 2 stations after c: \[ b + 2 \leq c \leq k \] Hence c can take on \( k - (b+2) + 1 = k - b - 1 \) values. Therefore the total number of possibilities of a,b,c is \[ \sum_{a=1}^{k-4} \sum_{b=a+2}^{k-2} (k - b - 1) \]

  11. JamesJ
    • 4 years ago
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    Evaluate that (slightly brutal, make a mistake, fix it, make another, start again, finally get it) you'll find it is equal to \[ \frac{1}{6} (k^3 - 9k^2 + 26k - 24) \] I'd love you to show this is equal to \[ {k-2 \choose 3} \]

  12. anonymous
    • 4 years ago
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    Sure, here is it: \[ {K-2 \choose 3} =\frac{ (K-2)(K-3)(K-4)}{6} = \frac{1}{6} (K^3 - 9K^2 + 26K - 24) \]

  13. JamesJ
    • 4 years ago
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    Exactly. Now, we know for sure that \[ { k-2 \choose 3} \] is right, we can say that our (my?) intuitive derivation of this is right: Given the placement of the first station, one is knocked out; then after the second is place, so is another. Hence we need to choose 3 stations from k-2. I would not have trusted that intuitive derivation without the more formal derivation. But it's nice to know it does indeed work.

  14. anonymous
    • 4 years ago
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    James what you are saying intuition is a standard trick in combinatorics, but this doesn't answer for the more general case of X stations from K intermediate stations.

  15. anonymous
    • 4 years ago
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    Btw I think I should now try to hunt down a pure combinatorial analysis of this solution.

  16. JamesJ
    • 4 years ago
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    Gee, you're so hard to please. Round of applause please. ===== Now, to the general problem, an obvious hypothesis is \[ { k-x+1 \choose x } \]

  17. anonymous
    • 4 years ago
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    Haha you did great job man :D Clap Clap Clap :D

  18. JamesJ
    • 4 years ago
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    I feel better now. I guess I don't trust that "intuitive derivation" because that knocking out of options feels like a process that could go wrong; it doesn't feel sufficiently well defined. Or it could just be I don't spend enough time around these sorts of problems. Or I'm just very skeptical, which I think is a good thing. ;-)

  19. JamesJ
    • 4 years ago
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    Anyway. If you find another derivation, let me know!

  20. anonymous
    • 4 years ago
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    Sure :)

  21. AravindG
    • 4 years ago
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    hey :)

  22. JamesJ
    • 4 years ago
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    So Aravind, I did all this work for you. Did you ever use this solution?

  23. AravindG
    • 4 years ago
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    hmm ...yes!

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