Question

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

Answer 1

NotHard

Answer 2

so its all combinations of (a, a+2, a+4)?

Answer 3

where a is an integer from 1 to 16?

Answer 4

which means 16 sets?

Answer 5

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}

Answer 6

yes - I think you are right there

Answer 7

actually that should be (20C3)-(ALL sets of 3 that has at least two consecutive numbers)

Answer 8

20*19*18

Answer 9

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

Answer 10

nvm, my logic's flawed

Answer 11

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.

Answer 12

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.

Answer 13

I think it would be:
(20-3)!*(20-4)!*(20-5)!*...*(20-19)!

Answer 14

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

Answer 15

\[\left(\begin{matrix}18 \\ 3\end{matrix}\right)\]