SOLUTION: Help! 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, b

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Question 1146066: Help!
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?
Answer by VFBundy(438) About Me  (Show Source):
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:

5 regular friends and 0 other friends
4 regular friends and 1 other friend

Ways to invite 5 regular friends and 0 other friends:

9C5 = 9%21%2F%285%21%2A4%21%29 = 126

Ways to invite 4 regular friends and 1 other friend:

9C4 * 2C1 = 9%21%2F%284%21%2A5%21%29 * 2%21%2F%281%21%2A1%21%29 = 126 * 2 = 252

So...there are a total of 126 + 252...or 378 ways she can do the invites.