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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?
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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asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1so its all combinations of (a, a+2, a+4)?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1where a is an integer from 1 to 16?

ChrisS
 4 years ago
Best ResponseYou've already chosen the best response.1well 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
 4 years ago
Best ResponseYou've already chosen the best response.1yes  I think you are right there

ChrisS
 4 years ago
Best ResponseYou've already chosen the best response.1actually that should be (20C3)(ALL sets of 3 that has at least two consecutive numbers)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.020 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
 4 years ago
Best ResponseYou've already chosen the best response.0nvm, my logic's flawed

ChrisS
 4 years ago
Best ResponseYou've already chosen the best response.1So 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
 4 years ago
Best ResponseYou've already chosen the best response.1I 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
 4 years ago
Best ResponseYou've already chosen the best response.1I think it would be: (203)!*(204)!*(205)!*...*(2019)!

ChrisS
 4 years ago
Best ResponseYou've already chosen the best response.1I have found the following formula \[\left(\begin{matrix}nk+1 \\ k\end{matrix}\right)\] Which in this case would give me

ChrisS
 4 years ago
Best ResponseYou've already chosen the best response.1\[\left(\begin{matrix}18 \\ 3\end{matrix}\right)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well, the answer's posted on the internet: http://www.math.ucsb.edu/~davie/page1/page6/page20/assets/Midterm%20Solutions.pdf question 4

ChrisS
 4 years ago
Best ResponseYou've already chosen the best response.1I'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
 4 years ago
Best ResponseYou've already chosen the best response.1ok... 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!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no prob, i just wanted to know what the answer was
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