document.write( "Question 728377: In how many ways can a mixed commitee of 5 be selected from 5 boys and 4 girls if the boys must be in the majority? \n" ); document.write( "
Algebra.Com's Answer #445388 by jim_thompson5910(35256)\"\" \"About 
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Case 1) If you have 5 boys on the committee, then you have 5 C 5 = 1 way to make this committee\r
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\n" ); document.write( "\n" ); document.write( "Case 2) If you have 4 boys and 1 girl on the committee, then you have (5 C 4)*(4 C 1) = (5)*(4) = 20 ways to do this.\r
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\n" ); document.write( "\n" ); document.write( "Case 3) If you have 3 boys and 2 girls on the committee, then you have (5 C 3)*(4 C 2) = (10)*(6) = 60 ways to do this.\r
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\n" ); document.write( "\n" ); document.write( "You stop here because you can't have 2 boys and 3 girls. The requirement is that the boys must be in the majority, so this is why we stop here.\r
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\n" ); document.write( "\n" ); document.write( "Add up all the results: 1+20+60 = 81\r
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\n" ); document.write( "\n" ); document.write( "So there are 81 ways to do this.\r
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\n" ); document.write( "\n" ); document.write( "Note: if you must have at least one girl on the committee, then there are 80 ways to do this (and you ignore case 1) \n" ); document.write( "
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