JamesJ
  • JamesJ
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?
Mathematics
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SOLVED
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katieb
  • katieb
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JamesJ
  • JamesJ
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.
JamesJ
  • JamesJ
Now how to generalize ...
anonymous
  • anonymous
James Please hold on sec, I want to give it a fair try :)

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JamesJ
  • JamesJ
Sure.
anonymous
  • anonymous
Guessing from pattern: \( \large \binom{K-2}{3} \)
anonymous
  • anonymous
Do you have a solution? @james
JamesJ
  • JamesJ
I do.
JamesJ
  • JamesJ
Intuitively, your hypothesis makes sense.
anonymous
  • anonymous
Okay, so what's your analysis ? :)
JamesJ
  • JamesJ
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 \[ \sum_{a=1}^{k-4} \sum_{b=a+2}^{k-2} (k - b - 1) \]
JamesJ
  • JamesJ
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 \[ {k-2 \choose 3} \]
anonymous
  • anonymous
Sure, here is it: \[ {K-2 \choose 3} =\frac{ (K-2)(K-3)(K-4)}{6} = \frac{1}{6} (K^3 - 9K^2 + 26K - 24) \]
JamesJ
  • JamesJ
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Btw I think I should now try to hunt down a pure combinatorial analysis of this solution.
JamesJ
  • JamesJ
Gee, you're so hard to please. Round of applause please. ===== Now, to the general problem, an obvious hypothesis is \[ { k-x+1 \choose x } \]
anonymous
  • anonymous
Haha you did great job man :D Clap Clap Clap :D
JamesJ
  • JamesJ
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. ;-)
JamesJ
  • JamesJ
Anyway. If you find another derivation, let me know!
anonymous
  • anonymous
Sure :)
AravindG
  • AravindG
hey :)
JamesJ
  • JamesJ
So Aravind, I did all this work for you. Did you ever use this solution?
AravindG
  • AravindG
hmm ...yes!

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