ChrisS
  • ChrisS
How many sets of three integers between 1 and 20 are possible if no two consecutive integers are to be in a set?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Hero
  • Hero
NotHard
asnaseer
  • asnaseer
so its all combinations of (a, a+2, a+4)?
asnaseer
  • asnaseer
where a is an integer from 1 to 16?

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asnaseer
  • asnaseer
which means 16 sets?
ChrisS
  • ChrisS
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}
asnaseer
  • asnaseer
yes - I think you are right there
ChrisS
  • ChrisS
actually that should be (20C3)-(ALL sets of 3 that has at least two consecutive numbers)
anonymous
  • anonymous
20*19*18
anonymous
  • anonymous
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
anonymous
  • anonymous
nvm, my logic's flawed
ChrisS
  • ChrisS
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.
ChrisS
  • ChrisS
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.
asnaseer
  • asnaseer
I think it would be: (20-3)!*(20-4)!*(20-5)!*...*(20-19)!
ChrisS
  • ChrisS
I have found the following formula \[\left(\begin{matrix}n-k+1 \\ k\end{matrix}\right)\] Which in this case would give me
ChrisS
  • ChrisS
\[\left(\begin{matrix}18 \\ 3\end{matrix}\right)\]
asnaseer
  • asnaseer
|dw:1327971872830:dw|
anonymous
  • anonymous
well, the answer's posted on the internet: http://www.math.ucsb.edu/~davie/page1/page6/page20/assets/Midterm%20Solutions.pdf question 4
ChrisS
  • ChrisS
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.
ChrisS
  • ChrisS
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!
asnaseer
  • asnaseer
yw
anonymous
  • anonymous
no prob, i just wanted to know what the answer was

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