document.write( "Question 1146066: Help!
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document.write( "A woman has 11 close friends. In how many ways can she invite 5 of them for dinner, if there are 2 who are not in good talking terms and so both can not be at the dinner, but at most one of them can be invited? \n" );
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Algebra.Com's Answer #767366 by VFBundy(438)![]() ![]() You can put this solution on YOUR website! There are 9 \"regular\" friends and 2 \"other\" friends. (\"Other\" friends are friends that cannot both be invited at the same time.) The woman can invite the following combinations: \n" ); document.write( " \n" ); document.write( "5 regular friends and 0 other friends \n" ); document.write( "4 regular friends and 1 other friend \n" ); document.write( " \n" ); document.write( "Ways to invite 5 regular friends and 0 other friends: \n" ); document.write( " \n" ); document.write( "9C5 = \n" ); document.write( " \n" ); document.write( "Ways to invite 4 regular friends and 1 other friend: \n" ); document.write( " \n" ); document.write( "9C4 * 2C1 = \n" ); document.write( " \n" ); document.write( "So...there are a total of 126 + 252...or 378 ways she can do the invites. |