A lady gives a dinner party for six guests. The number of ways in which they may be selected from ten friends will not attend the party together, is?

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112

140

164

None of these

answer is B.

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Detailed Solution

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.

C_{6}^{8} = 28

Therefore, there are 28 ways to invite 6 friends when the two friends will not be going to the party. 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

C_{5}^{8}

. Similarly, if the other friend decides to attend the party, the number of ways to invite the other 5 guests is

C_{5}^{8}

.
Therefore, the number of ways to invite the guests if either, if the two friends is going to the party, is

{2 \times C}_{5}^{8}

= 112
Therefore, the total number of ways to invite six guests out of ten with the given condition is 28+112 =140.
Hence, Option (2) is the correct answer.

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