# Am I missing something?

I know this question looks simple and might in fact be simple, but, please help me with this one:- If $$n$$ things are arranged in a circular order, then show that the number of ways of selecting four of the things no two of which are consecutive is $$\dfrac{n(n - 5)(n - 6)(n - 7)}{4!}$$.

My approach is that since, the things are arranged in circular order, lets select 4 thing such that there is $$a,b,c,d$$ things between each of them where $$a,b,c,d$$ are positive integers and satisfy the equation $$a + b + c + d = n-4$$ which can be obtained by $$\dbinom{n-4-1}{4-1} = \dfrac{(n-5)(n-6)(n-7)}{3!}$$ but the required answer is $$\dfrac{n}{4}$$ times this answer. Where am I going wrong, a detailed solution is appreciated.

Note by Ashish Siva
2 years ago

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Hi @Ashish Siva The first answer is the correct one. Without going into too much math here, I think your approach is correct only if you fix one of the objects as a selected one. In reality all $$n$$ objects can be either selected or not. So, this will be a subset of the complete solution.

For example if you have $$9$$ objects, there are $$9$$ ways to select the objects. (consistent with the first answer). However, if you pick one as selected, then there are only $$4$$ ways to pick the objects (consistent with your answer).

Make sense?

- 2 years ago

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Ya, thanks! Surely makes sense!

- 2 years ago

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