document.write( "Question 607661: How many ways can a committee of 2 boys and 2 girls and an alternate of either gender be chosen from 5 boys and 6 girls? \n" ); document.write( "
Algebra.Com's Answer #382838 by Edwin McCravy(20055)\"\" \"About 
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document.write( "We can choose the two boys any of 5C2 or 10 ways.\r\n" );
document.write( "For each of the 10 ways to choose the boys, we can\r\n" );
document.write( "choose the two girls any of 6C2 or 15 ways.  \r\n" );
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document.write( "That's 10·15 or 150 ways to choose the two boys and \r\n" );
document.write( "the two girls.  \r\n" );
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document.write( "For each of those 150 ways which we can choose the two \r\n" );
document.write( "boys and the two girls, we can choose the alternate as \r\n" );
document.write( "any of the 7 remaining people.  That's 150·7 or 1050 \r\n" );
document.write( "ways.\r\n" );
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document.write( "Answer: 5C2·6C2·7 = 10·15·7 = 1050 ways.\r\n" );
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document.write( "Edwin
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