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.

#Combinatorics

Easy Math Editor

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Comments

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?

Geoff Pilling - 4 years, 10 months ago

Ya, thanks! Surely makes sense!

Ashish Menon - 4 years, 10 months ago

@Geoff Pilling @Hung Woei Neoh @Sambhrant Sachan please do comment.

Ashish Menon - 4 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$