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In essence, how many unique ways can I choose a subset of N people such that there exist 3 people in the subset who are adjacent in the original set.

e.g. N = 4 Lets label the people {1,2,3,4} I can choose {1,2,3} , {2,3,4} , { 1,2,3,4} making a total of 3 ways.

Here I assume that {1,2,3} = {3,2,1} etc. are the same

Is there any general formula I can derive for calculating this?

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It seems this question has been picked up from currently running contest, which is against rules! Admin Please verify the same. I don't have enough reputation to comment otherwise I would have done that. :( – rajneesh2k10 Sep 10 '12 at 16:35
Yes, it is based on the above mentioned problem. I was not aware that this was against math.stackexchange rules. I was just trying to find a simple way to solve this. – varrunr Sep 10 '12 at 20:45

Let us call the number of ways of choosing a subset of N people such that there exist 3 people in the subset who are adjacent in the original set as $a_n$. Call such subsets good.

Let's try to find a recurrence for $a_n$.

Now, let us take any such subset of $N-1$ people, and consider that set with $N$ in it and not in it. Clearly both are distinct good subsets. So, this amounts to $2 a_{n-1}$ ways.

So, the elements not yet considered have $N$ in the triplet, and in fact the triplet $N-2, N-1,N$ should be the only one inside the subset, otherwise it has been calculated before. So, this amounts to $2^{n-4} - a_{n-4}$ ways, as we need the number of subsets of $N-4$ people not having any such triplet (Note that the element $N-3$ cannot be in the set).

So, we get the recurrence as $$a_{n} = 2a_{n-1} + 2^{n-4} - a_{n-4}$$ with initial conditions as $a_1=a_2=0, a_3=1, a_4=3$.

As regards to a general formula, the use of OEIS leads to this which tells us that the generating function is $$G(x) = \frac{x^3}{(1-2x)(1-x-x^2-x^3)}$$ and also gives a nicer looking recurrence: $$a_n = 3a_{n-1} - a_{n-2} - a_{n-3} - 2a_{n-4}$$

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I am actually looking for an expression involving N and constants. – varrunr Sep 4 '12 at 22:48
var, there is a standard way to go from that last recurrence to the kind of formula you want - do you know how to do this? Bear in mind that Rijul seems to be using $n$ and $N$ interchangeably. If you don't know how to solve constant-coefficient linear recurrences, follow Rijul's link to OEIS where there's a formula in terms of tribonacci, and then follow the link to a formula for tribonacci. But it really pays to learn how to solve these recurrences. – Gerry Myerson Sep 5 '12 at 2:09
@GerryMyerson Yes, I can solve recurrences. I was just looking to see if there is a more straightforward solution to get a relationship with N , rather than creating a recurrence and then solving it to get N. – varrunr Sep 5 '12 at 4:12
varrunr, when you get such a 4 term linear recurrence, the general formula in terms of n is often nasty and pretty difficult to work with, for example, I would rather work with the Fibonacci Recurrence than the Binet's Formula if I can avoid it. Anyway, I don't know how to get a straightforward solution to reach the Binet's Formula, which doesn't involve all this machinery. – Rijul Saini Sep 5 '12 at 6:19

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