document.write( "Question 1061558: In how many ways can 5 men and 7 women make up a special
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document.write( "committee if 5 persons are selected and at least 3 committee
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document.write( "members must be women? \n" );
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Algebra.Com's Answer #676325 by Edwin McCravy(20055)![]() ![]() You can put this solution on YOUR website! In how many ways can 5 men and 7 women make up a special \n" ); document.write( "committee if 5 persons are selected and at least 3 committee \n" ); document.write( "members must be women? \n" ); document.write( " \r\n" ); document.write( "There can be either exactly 3 women, exactly 4 women, \r\n" ); document.write( "or exactly 5 women.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "A. Number of ways he can have 3 women and 2 men:\r\n" ); document.write( "\r\n" ); document.write( "Choose the 3 women 7C3 ways and the 2 men 5C2 ways.\r\n" ); document.write( "That's (7C3)(5C2) = (35)(10) = 350 ways.\r\n" ); document.write( "\r\n" ); document.write( "B. Number of ways he can have 4 women and 1 man:\r\n" ); document.write( "\r\n" ); document.write( "Choose the 4 women 7C4 ways and the 1 man 5C1 ways.\r\n" ); document.write( "That's (7C4)(5C1) = (35)(5) = 175 ways.\r\n" ); document.write( "\r\n" ); document.write( "C. Number of ways he can have 5 women and 0 men:\r\n" ); document.write( "\r\n" ); document.write( "Choose the 5 women 7C5 ways and the 0 men 5C0 ways.\r\n" ); document.write( "That's (7C5)(5C0) = (21)(1) = 21 ways.\r\n" ); document.write( "\r\n" ); document.write( "Total = 350+175+21 = 546 ways.\r\n" ); document.write( "\r\n" ); document.write( "Edwin\n" ); document.write( " |