document.write( "Question 1198404: In hw many ways can three items be selected from a group of six items. Use letters A B C D E and F to identity the items and list each of the different combination of three items \n" ); document.write( "
Algebra.Com's Answer #831995 by Edwin McCravy(20060)\"\" \"About 
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document.write( "This is the number of combinations of 6 things taken 3 at a time.\r\n" );
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document.write( "{A,B,C} {A,B,D} {A,B,E} {A,B,F} {A,C,D} \r\n" );
document.write( "{A,C,E} {A,C,F} {A,D,E} {A,D,F} {A,E,F}\r\n" );
document.write( "{B,C,D} {B,C,E} {B,C,F} {B,D,E} {B,D,F} \r\n" );
document.write( "{B,E,F} {C,D,E} {C,D,F} {C,E,F} {D,E,F} \r\n" );
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document.write( "20 ways. The formula is \"6C3\"\"%22%22=%22%22\"\"%286%2A5%2A4%29%2F%283%2A2%2A1%29\"\"%22%22=%22%22\"\"120%2F6\"\"%22%22=%22%22\"\"20\"\r\n" );
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document.write( "To form the numerator, the first factor is the number we have to select from,\r\n" );
document.write( "then the next factor is 1 less, and the next factor is 1 less than that, etc. if\r\n" );
document.write( "necessary.  We stop when the number of factors is equal to the number we are\r\n" );
document.write( "choosing.\r\n" );
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document.write( "To form the denominator, the first factor is the number we are choosing, and\r\n" );
document.write( "the next factor is 1 less, and the next factor is 1 less, etc. if necessary. We\r\n" );
document.write( "stop when we get to 1.  There should be the same number of factors in the\r\n" );
document.write( "denominator as in the denominator.  \r\n" );
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document.write( "Form the fraction, then simplify to an integer.\r\n" );
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document.write( "Edwin
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