document.write( "Question 1149685: Nine players are to be divided into two teams of four and one umpire. If two
\n" ); document.write( "particular people cannot be on the same team, how many different combinations
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Algebra.Com's Answer #771188 by Edwin McCravy(20055) $\"\"$ $\"About$

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Suppose the nine players are A,B,C,D,E,F,G,H,I\r\n" );
document.write( "Suppose A and B cannot be on the same team.  There are two cases.\r\n" );
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document.write( "Case 1.  Neither A nor B is the umpire. \r\n" );
document.write( "A is on one team and B is on the other.  Then we choose the umpire from the\r\n" );
document.write( "7 players C,D,E,F,G,H,I.  That's 7 ways to pick the umpire.  \r\n" );
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document.write( "We have 6 players left. We choose 3 to play on the team with A in 6C3 = 20\r\n" );
document.write( "ways, and the remaining 3 will play on the team with B in 3C3 = 1 way.\r\n" );
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document.write( "That's 7∙20∙1 = 140 ways for case 1.\r\n" );
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document.write( "Case 2.  A or B is the umpire.\r\n" );
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document.write( "We choose the umpire 2 ways.\r\n" );
document.write( "Then we have 8 players choose 4 on one team in 8C4 = 70 ways, and that\r\n" );
document.write( "leaves 4C4 = 1 way to put the others on the other team.\r\n" );
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document.write( "That's 2∙70 = 140 ways for case 2.\r\n" );
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document.write( "So that's 140 for case 1 and 140 for case 2, a total of 280 ways. \r\n" );
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document.write( "Edwin

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