Question

A lady gives a dinner party for six guests. The number of ways in which they may be selected from ten friends, if two of the friends will not attend the party together, is?
(a) 112
(b) 140
(c) 164
(d) None of these

Answer 1

Hint: We will consider two cases. The first case will be when both friends will not be attending the party. So we will choose 6 out of the 8 remaining friends. The second case will be when either of the two friends will be coming to the party. So, we will have to choose the other 5 guests from the remaining 8 friends.

Complete step by step answer:
The lady has ten friends. Two of these friends will not agree to go to the party together. So, let us consider the following case: neither of the two friends will attend the party. So, the number of ways six guests can be invited in this case will be equal to the number of ways to choose 6 friends out of the remaining 8 friends. So, we have
 $ \begin{align}
  & {}^{8}{{C}_{6}}=\dfrac{8!}{6!2!} \\
 & \Rightarrow {}^{8}{{C}_{6}}=\dfrac{8\times 7}{2} \\
 & \therefore {}^{8}{{C}_{6}}=28 \\
\end{align} $
Therefore, there are 28 ways to invite 6 friends when the two friends will not be going to the party.
Now, we will consider the next case, which is that one of the two friends decides to attend the party. So, if either of the two decide to attend the party, then the rest 5 guests will be chosen from the remaining 8 friends. Therefore, if one friend decides to attend the party, the number of ways to invite the other 5 guests is $ {}^{8}{{C}_{5}} $ . Similarly, if the other friend decides to attend the party, the number of ways to invite the other 5 guests is $ {}^{8}{{C}_{5}} $ . Therefore, the number of ways to invite the guests if either, if the two friends is going to the party, is the following,
 $ \begin{align}
  & {}^{8}{{C}_{5}}+{}^{8}{{C}_{5}}=2\times {}^{8}{{C}_{5}} \\
 & \Rightarrow 2\times {}^{8}{{C}_{5}}=2\times \dfrac{8!}{5!3!} \\
 & \Rightarrow 2\times {}^{8}{{C}_{5}}=2\times \dfrac{8\times 7\times 6}{3\times 2} \\
 & \therefore 2\times {}^{8}{{C}_{5}}=112 \\
\end{align} $
Therefore, the total number of ways to invite six guests out of ten with the given conditions is $ 28+112=140 $ .
 Hence, the correct option is (b).


Note:
 It is important to classify the cases for such type of questions. This helps us in understanding the way in which we should count the possibilities occurring. We used combinations for counting the possibilities in these questions. It is useful to know the concept of permutations and combinations for such type of questions.