# Thread: Calculating number of non consecutive numbers

1. ## Calculating number of non consecutive numbers

Hi,

Can someone help me with a formula that will give me the number of non consecutive numbers in a given number of events.

So, if I had thirty selections how can I work out how many combinations of three numbers there are in the thirty selections. ie. 1,3,5 1,3,6 1,3,7 etc

I would like a formula that allows me to put in the total number of events coupled with the number of selections. So, I would like to be able to work out, say, the number of four non consecutive groups in a total of, say, ten selections.

As you've probably realised this is linked to some gambling probabilities. Hope you can help.

parkingticket60

2. I am not sure that I understand what you have written.
Here is a stab at an answer.
Consider the set of digits: $\{0,1,2,3,4,5,6,7,8,9\}$.
I can select four of those with no two consecutive in $\dbinom{7}{4}=\dfrac{7!}{4!\cdot 3!}=35$ ways.
We can select up to five digits with no consecutive digits.

If you had 15 items you can select up to 8 with no two consecutive.
For n items you can select up to $\left\lceil {\frac{n}{2}} \right\rceil$ with no two consecutive. That is the ceiling function.
The number of ways of selecting k items is $\dbinom{n-k+1}{k}$.

If that is not what mean, please try to make is clear what you do mean.

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# how to find the non conseutive numbers

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