document.write( "Question 1146066: Help!
\n" ); 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?
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Algebra.Com's Answer #767366 by VFBundy(438)\"\" \"About 
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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:
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\n" ); document.write( "5 regular friends and 0 other friends
\n" ); document.write( "4 regular friends and 1 other friend
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\n" ); document.write( "Ways to invite 5 regular friends and 0 other friends:
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\n" ); document.write( "9C5 = \"9%21%2F%285%21%2A4%21%29\" = 126
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\n" ); document.write( "Ways to invite 4 regular friends and 1 other friend:
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\n" ); document.write( "9C4 * 2C1 = \"9%21%2F%284%21%2A5%21%29\" * \"2%21%2F%281%21%2A1%21%29\" = 126 * 2 = 252
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\n" ); document.write( "So...there are a total of 126 + 252...or 378 ways she can do the invites.
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