Re: Number of combinations

Quote:

Originally Posted by

**billobillo** If I have numbers from 1 to N and I want to select K out of them.

In how many ways I can select these K numbers such that they contain at least two consecutive numbers.

For example if N=7 and K=3, I want to select combinations such as {1,2,3) , {2,3,7} , {3,5,6} but not {1,3,5} or {2,4,7} ....

First note that if $\displaystyle K \leqslant \left\lceil {\frac{N}{2}} \right\rceil $, that is the ceiling function, there are $\displaystyle \binom{N-K+1}{K}$ of pick a subset of $\displaystyle K$ numbers with __no__ consecutive numbers. So subtract from the total.

If $\displaystyle K > \left\lceil {\frac{N}{2}} \right\rceil $ then any subset of $\displaystyle K$ numbers with **must contain consecutive numbers.**

I do not understand the second part at all.

Re: Number of combinations

Quote:

Originally Posted by

**Plato** First note that if $\displaystyle K \leqslant \left\lceil {\frac{N}{2}} \right\rceil $, that is the ceiling function, there are $\displaystyle \binom{N-K+1}{K}$ of pick a subset of $\displaystyle K$ numbers with __no__ consecutive numbers. So subtract from the total.

If $\displaystyle K > \left\lceil {\frac{N}{2}} \right\rceil $ then any subset of $\displaystyle K$ numbers with **must contain consecutive numbers.**

I do not understand the second part at all.

Thank you for the reply, I guess you are the genius of this forum :), in first part the difference between any 2 of the K numbers should be >= "1"; you answer is correct.

In the second part , I need to choose a number other than "1" i.e. if D="2" , then the difference between any 2 of the K numbers should be >= "2"

For example : If D=2, I need set like {1,3,5} where 3-1>=2 and 5-3>=2 and 5-1>=2 OR set like {2,4,7} ......

Or If D=3 I need number of sets like {1,4,7} where 4-1>=3 and 7-4>=3 and 7-1>=3.

Is there a way to create a general formula that takes N,K and D and give me the number of combinations allowed?

Regards.