Question

A lady gives a dinner party for five guests to be selected from nine friends. the number of ways of forming the party of 5, given that two particular friends a and b will not attend the party together is

Answer 1

To make it easier to understand we will note the friends as A B C D E F G H I. A&B don't come together : Case 1 : Neither one of them comes (Selecting 5 guests from remaining 7 guests) Case 2 : Only one of them comes (Selecting 4 guests from remaining 7 guests and one from A&B). So, (7x6) / (2x1) = 21 possible ways and (2x1) x (7x6x5) / (3x2x1) = 70 possible ways and at last you just have to add 21 and 70 and you get 91 possible ways in total so that's your answer.

Hope I helped :)