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Possible Duplicate:
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.

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rspr
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  • Are you sure it's $20$ for $n=5$? I think it should be $6$. – Quixotic Sep 07 '12 at 04:58
  • $ \frac{1}{2} \left(N^2-3 N+2\right)$ – Quixotic Sep 07 '12 at 05:02
  • 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 $8$. – Brian M. Scott Sep 07 '12 at 05:04
  • srry for confusion its 8 for 5 and so on(typo mistkae), I have updated the question – rspr Sep 07 '12 at 06:09
  • 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) – rspr Sep 07 '12 at 06:14
  • @Marc Ah, I see. That is not how I had interpreted the question. When I read "choose 3 and ore consecutive numbers," I interpret this as the number of ways of choosing 3 consecutive, or 4 consecutive, or 5, etc. But in the example he posted, I see this is not the case. – davidlowryduda Sep 09 '12 at 19:36

1 Answers1

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HINT:

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

Quixotic
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