SOLUTION: Nine players are to be divided into two teams of four and one umpire. If two particular people cannot be on the same team, how many different combinations are possible?

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Question 1149685: Nine players are to be divided into two teams of four and one umpire. If two
particular people cannot be on the same team, how many different combinations
are possible?

Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
Suppose the nine players are A,B,C,D,E,F,G,H,I
Suppose A and B cannot be on the same team.  There are two cases.

Case 1.  Neither A nor B is the umpire. 
A is on one team and B is on the other.  Then we choose the umpire from the
7 players C,D,E,F,G,H,I.  That's 7 ways to pick the umpire.  

We have 6 players left. We choose 3 to play on the team with A in 6C3 = 20
ways, and the remaining 3 will play on the team with B in 3C3 = 1 way.

That's 7∙20∙1 = 140 ways for case 1.

Case 2.  A or B is the umpire.

We choose the umpire 2 ways.
Then we have 8 players choose 4 on one team in 8C4 = 70 ways, and that
leaves 4C4 = 1 way to put the others on the other team.

That's 2∙70 = 140 ways for case 2.

So that's 140 for case 1 and 140 for case 2, a total of 280 ways. 

Edwin

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