document.write( "Question 1019009: a committee of 5 people must be selected from 5 men and 8 women. 13 people in ALL.\r
\n" ); document.write( "\n" ); document.write( "How many ways can a selection be done if there are at least 3 women on the committee?
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Algebra.Com's Answer #634988 by mathmate(429)\"\" \"About 
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\n" ); document.write( "Question:
\n" ); document.write( "a committee of 5 people must be selected from 5 men and 8 women. 13 people in ALL.
\n" ); document.write( "How many ways can a selection be done if there are at least 3 women on the committee?
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\n" ); document.write( "Solution:
\n" ); document.write( "There are three cases:
\n" ); document.write( "1. 3 women and two men
\n" ); document.write( "C(5,2)*C(8,3)=560 ways
\n" ); document.write( "2. 4 women and 1 man:
\n" ); document.write( "C(5,1)*C(8,4)=350 ways
\n" ); document.write( "3. 5 women and no men:
\n" ); document.write( "C(5,0)*C(8,5)=56 ways
\n" ); document.write( "Total=560+350+56=966 ways.
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\n" ); document.write( "Following is the incorrect way to solve the problem:
\n" ); document.write( "Choose 3 women and two more from the rest:
\n" ); document.write( "C(8,3)*C(10,2)=2520 >>966
\n" ); document.write( "This way, we're overcounting, since some of the women chosen out of the remaining 10 could have been counted already, for example, ABC were chosen first, and G was chosen next. But then it would be the same if ABG was chosen first, and C was chosen later.
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