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.

## Comments

Sort by:

TopNewestHi @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? – Geoff Pilling · 9 months, 4 weeks ago

Log in to reply

– Ashish Siva · 9 months, 3 weeks ago

Ya, thanks! Surely makes sense!Log in to reply

@Geoff Pilling @Hung Woei Neoh @Sambhrant Sachan please do comment. – Ashish Siva · 9 months, 4 weeks ago

Log in to reply