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Counting subsets containing three consecutive elements (previously Summation over large values of nCr)

Suppose we have a set like (1,2,3) then there is only one way to choose 3 consecutive number...its (1,2,3)....for a sets of 4 (1,2,3,4) we have 3 ways ( (1,2,3), (2,3,4), (1,2,3,4)) for five its 8 ,for 6 its 20, for 7 its 47 and so on....So for a given N, I can get the answer by applying brute force, and calculating all such subset having 3 or more consecutive number. Here I am just trying to find out a pattern, a technique to get the number of all such subset for a given N. The problem is further generalized to .....discover m consecutive number within a set of size N.

  • $\begingroup$ Are you sure it's $20$ for $n=5$? I think it should be $6$. $\endgroup$ – Quixotic Sep 7 '12 at 4:58
  • $\begingroup$ $ \frac{1}{2} \left(N^2-3 N+2\right)$ $\endgroup$ – Quixotic Sep 7 '12 at 5:02
  • $\begingroup$ Do all of the numbers in the subset have to be consecutive, or does the subset just have to contain at least three consecutive numbers? In other words, is $\{1,2,3,5\}$ acceptable? If all of them have to be consecutive, see the hints below; if not, see this answer. In either case the number for $5$ is not $20$; it’s either $6$ or $8$. $\endgroup$ – Brian M. Scott Sep 7 '12 at 5:04
  • $\begingroup$ srry for confusion its 8 for 5 and so on(typo mistkae), I have updated the question $\endgroup$ – rspr Sep 7 '12 at 6:09
  • $\begingroup$ for 5 the 8's are as...(3,4,5), (2,3,4), (1,2,3,4), (1,3,4,5), (1,2,3), (1,2,3,5), (1,2,3,4), (1,2,3,4,5) $\endgroup$ – rspr Sep 7 '12 at 6:14


The number of selections of $k$ consecutive things out of $n$ things in a row is given by $n-k+1$.


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