8!/2!2!2!2!4! = 105

8 groups divided into four teams can be done in 8!/2!2!2!2!, but as order doesnt matter....divide by 4!

praky_rules wrote:8!/2!2!2!2!4! = 105

8 groups divided into four teams can be done in 8!/2!2!2!2!, but as order doesnt matter....divide by 4!

Can you please explain in greater detail?

Thanks!

there's a really good explanation by Ian here:
https://www.beatthegmat.com/friends-want ... 18991.html

praky_rules wrote:8!/2!2!2!2!4! = 105

8 groups divided into four teams can be done in 8!/2!2!2!2!, but as order doesnt matter....divide by 4!

Could you explain further ? sorry for the trouble.

Ok. Assume that there are 4 teams(named A,B,C and D).
choose 2 from 8 for Team A in 8C2 ways.
choose 2 from 6 for Team B in 6C2 ways.
choose 2 from 4 for Team B in 4C2 ways.
The remaining two automatically go to D.

So the total number of ways you can assign 8 tennis pros into 4 teams A,B,C and D = 8C2*6C2*4C2 = 8!/(2!)^4.

But the names A,B,C,D do not matter!!...they are interchangeable....that means whether you name a team A or B or C or D it does not matter as all the teams have only 2 tennis champs!!....so divide the total by recounts = (4!)(= number of ways 4 teams can be named A,B,C,D.)...

I hope this helps!
Thanks