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I have an interesting question which goes as follows:

The number of ways in which a host lady can invite for a party of 8 out of 12 people of whom two do not want to attend the party together is.

$(a) 2 × C(11,7) + C(10,8)$ $(b) C(10,8) + C(11,7)$ $(c) C(12,8) - C(10,6)$

Now I know one of the methods to solve it.

Say the two friends are A and B. The number of ways A can attend the party when B isn't = C(10,7) = number of ways you can choose 7 people out of 10, by fixing A to attend the party and B to not attend the party. The same follows for B.

So the number of ways either A or B can attend the party = 2 × C(10,7) Now we add in this, the number of ways when people can attend the party, when both A and B don't attend i.e. C(10,8) = choosing all 8 people out of 10 (leaving A and B)

This tells me that the first option is slightly varying from the answer.

The second option is kinda same as first, only it does not consider B attending the party when A isn't, or vice-versa.

Coming to option three. It seems to say that you find all the combinations in which people can attend the party and subtract something from it to get the answer. What my question is, I'm not getting what that "something" is and where does it come from? (Somethin' = C(10,6))

user231094
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1 Answers1

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The third option applies when both $A$ and $B$ attend the party. We have $10$ people left and we need to choose $6$ of them to complete a party of $8$.

Vasili
  • 10,690