SOLUTION: A woman has nine close friends. In how many ways can she invite exactly six of these friends to dinner, under the condition that two of her friends are not speaking and will not at

Question 252360: A woman has nine close friends. In how many ways can she invite exactly six of these friends to dinner, under the condition that two of her friends are not speaking and will not attend together?
a. 42 b. 49 c. 63 d. 72 e. 82
Answer by palanisamy(496) (Show Source): You can put this solution on YOUR website!
A woman has 9 friends.
Two of them do not talk to each other and will not attend the party together.
Let us call them as A and B.
Out of 9 friends 6 has to be selected for the dinner with the condition that if A is selected then B is not selected and if B is selected then A is not selected.
First we will include A and exclude B.
After selecting A, we have to select 5 from the remaing 7( exclude B) in 7C5 ways.
Similarly,After selecting B, we have to select 5 from the remaing 7( exclude A) in 7C5 ways.
Therefore total number of ways = 7C5 +7C5
= 2x7C5
= (2x7x6)/1x2
=42 ways

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