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## Homework Statement

If n things are arranged in circular order, then show that the number of ways of selecting four of the things no two of which are consecutive is

$$\frac{n(n-5)(n-6)(n-7)}{4!}$$

## Homework Equations

## The Attempt at a Solution

My approach to the problem is that I first select the objects without any restriction and subtract those cases when 4 things are consecutive, 3 things are consecutive and 2 things are consecutive. If this isn't the correct approach, please ignore what I have written below and suggest the alternative.

Number of ways to select 4 things without any restriction: ##\binom{n}{4}##

There are n ways for four things to be consecutive.

There are n(n-5) ways for 3 consecutive things. My logic behind this is that I first count the ways for 3 consecutive things and the last thing to selected must be from n-5 remaining things.

Now I feel that the trouble is caused due to the last case when two things are consecutive.

There are n ways to select groups of two consecutive things. Now the other two must be from n-4 things. Hence, ##n\binom{n-4}{2}## ways.

I have found the error in the last case but I am unable to exclude the duplicate selections. For example, I arrange 9 things represented by letters A,B,C,D,E,F,G,H and I respectively. Lets select AB and then EF. When we later on select EF and then AB, this selection is same. Likewise, there are many similar selection. I have no idea about removing these duplicate selections.

I hope I used the correct words.

Any help is appreciated. Thanks!