Question

Find the number of ways in which a lady can invite 6 guests selected from 10 friends so that two of the friends will not attend the party together ?.

Answer 1

Let us consider that the two particular friends are A and B

If A is invited among six guests and B is not then, number of combinations to select 5 more guests from the remaining 8 friends will be

${}^8{C}_5=\frac{8!}{5!3!}=56$

If B is invited among the six guests and A is not then, number of ways of selecting the remaining guest will be

${}^8{C}_5=\frac{8!}{5!3!}=56$

Suppose both A and B are not included in the six guests then the number of such combinations will be

${}^8{C}_6=\frac{8!}{6!2!}=28$

Therefore total number of sets of guests that can be selected $=56+56+28=140$.