Explain like I'm 5: 'overcounting' when selecting groups. (1 Viewer)

[emilios]

This has kind of become reflexive for me at this point but I don't quite get why I'm doing it. Say I have 11 people and I want to make 2 soccer teams consisting of 5 people with the remaining person designated as a referee. The total number of ways of doing this is:

11C5*6C5 / 2

Why do I divide by two? The textbook says its because the teams are arbitrary, but even that doesn't make sense to me. Am I dividing by 2 or by 2! ??

 

[Fade1233]

This has kind of become reflexive for me at this point but I don't quite get why I'm doing it. Say I have 11 people and I want to make 2 soccer teams consisting of 5 people with the remaining person designated as a referee. The total number of ways of doing this is:

11C5*6C5 / 2

Why do I divide by two? The textbook says its because the teams are arbitrary, but even that doesn't make sense to me. Am I dividing by 2 or by 2! ??

Yep thats right. If it was three teams then you would divide by 3!.

 

[emilios]

What about in this scenario;

Show that the no of ways 3n students can be divided into 3 groups consisting of n, n-1, n+1 students is (3n)!/[n!(n-1)! (n+1)!]

By the previous method, I did this

3nCn * 2nC(n-1) * (n+1)C(n+1) / (3!)

Which leads to an incorrect answer, in that I shouldn't have divided by 3!. Why is that?

 

[RealiseNothing]

You choose 5 people for team 1 then another 5 for team 2. However it is possible the 5 you choose for team 1 could be chosen as the 5 you have for team 2. You must divide by 2 to prevent yourself double counting this case.

 

[emilios]

Oh ok, thanks Realise, repped