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ChrisS

  • 4 years ago

How many sets of three integers between 1 and 20 are possible if no two consecutive integers are to be in a set?

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  1. Hero
    • 4 years ago
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    NotHard

  2. asnaseer
    • 4 years ago
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    so its all combinations of (a, a+2, a+4)?

  3. asnaseer
    • 4 years ago
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    where a is an integer from 1 to 16?

  4. asnaseer
    • 4 years ago
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    which means 16 sets?

  5. ChrisS
    • 4 years ago
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    well I think it's going to be (20C3)-(any set of three that has consecutive numbers) I'm just not sure how to compute that second set. I don't think (a,a+2,a+4) gets it for me because a valid set with the given restriction would be {1,10,18}

  6. asnaseer
    • 4 years ago
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    yes - I think you are right there

  7. ChrisS
    • 4 years ago
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    actually that should be (20C3)-(ALL sets of 3 that has at least two consecutive numbers)

  8. anonymous
    • 4 years ago
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    20*19*18

  9. anonymous
    • 4 years ago
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    20 possibilities for the first, 19 for the second since it can't be consecutive to the first, 18 possibilities for the third since it can't be consecutive to first or second integer

  10. anonymous
    • 4 years ago
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    nvm, my logic's flawed

  11. ChrisS
    • 4 years ago
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    So if the first number I chose happened to be 1 or 20, then that should leave me 18 possible choices for the second number. If it were any other number though, it would leave me 17 because I would have to exclude the number in front of and behind it.

  12. ChrisS
    • 4 years ago
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    I don't think I can repeat a number because I'm pulling sets. I believe an element can only be in a set once. If it were a list, I could have repeats.

  13. asnaseer
    • 4 years ago
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    I think it would be: (20-3)!*(20-4)!*(20-5)!*...*(20-19)!

  14. ChrisS
    • 4 years ago
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    I have found the following formula \[\left(\begin{matrix}n-k+1 \\ k\end{matrix}\right)\] Which in this case would give me

  15. ChrisS
    • 4 years ago
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    \[\left(\begin{matrix}18 \\ 3\end{matrix}\right)\]

  16. asnaseer
    • 4 years ago
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    |dw:1327971872830:dw|

  17. anonymous
    • 4 years ago
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    well, the answer's posted on the internet: http://www.math.ucsb.edu/~davie/page1/page6/page20/assets/Midterm%20Solutions.pdf question 4

  18. ChrisS
    • 4 years ago
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    I'm not quite following that, but if I am reading it right, that looks like it would give me a much higher number than I am looking for. With no restrictions, I would be using (20C3) which gives me 1140 subsets of three integers from the set {1...20}. Using this formula I found, I get (18C3) which generates 816 subsets. That number seems pretty reasonable.

  19. ChrisS
    • 4 years ago
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    ok... that one is (18C15) but (18C15) and (18C3) both give the same result... so I think I have the answer :) Thank you all for helping me work through this!

  20. asnaseer
    • 4 years ago
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    yw

  21. anonymous
    • 4 years ago
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    no prob, i just wanted to know what the answer was

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