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Three different numbers are chosen from 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. How many ways can the numbers be chosen so that no 2 of the 3 numbers are consecutive?
I'm completely stumped when it comes to this question. How do you deal with this kind of restriction? BTW, this is a combinations question (order of numbers picked does not matter). Thanks!
You should try an indirect method.
From the total of possible choices: pick 3 numbers:10C3
subtract the number of choices that you dont want, that would be:
all 3 numbers consecutive:8 possibilities
2 of the 3 numbers are consecutive:there are 9 posibilities to have 2 consecutive numbers, so there are 9*8posibilities for this case.
So, 10C3-8-72=40
The answer of 40 is incorrect. The error came from subtracting the same configuration more than once. That is an error of over-counting.
There is a well-known formula connected with this problem.
If S is a ordered set with n elements and k is a positive integer at most ceil(n/2), then the number of ways to choose a k-element subset of S having no consecutive members is Combin(n-k+1,k).
Thanks. I see what you mean.
2 of the 3 numbers are consecutive:there are 9 posibilities to have 2 consecutive numbers, so there are 9*8posibilities for this case include and some subset that have all 3 number consecutive.
Sorry