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by heshamelaziry » Sat Aug 29, 2009 2:55 pm
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105


Thank you. Please include an explanation.
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by praky_rules » Sat Aug 29, 2009 3:55 pm
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!
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by pathaniaus » Sat Aug 29, 2009 5:51 pm
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!
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by capnx » Sat Aug 29, 2009 6:31 pm
there's a really good explanation by Ian here:
https://www.beatthegmat.com/friends-want ... 18991.html
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by heshamelaziry » Sat Aug 29, 2009 6:43 pm
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.
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by praky_rules » Sun Aug 30, 2009 8:19 am
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
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