A lady gives dinner party to 5 guests to be selected from 9 friends. The…

Concept: When choosing r items from n items with the restriction that two particular items, say A and B, must not both be selected together, the sum rule gives the valid count as (selections with neither A nor B) + (selections with exactly one of A or B). This is equivalent to (all selections) − (selections containing both A and B together).

1. Let the two friends who will not attend together be A and B, so the remaining pool has 7 friends.

2. Case 1 (neither A nor B invited): choose all 5 guests from the remaining 7 friends: 7C5 = 21 ways.

3. Case 2 (exactly one of A or B invited): choose which one attends in 2C1 = 2 ways, then choose the remaining 4 guests from the other 7 friends: 7C4 = 35 ways, giving 2 × 35 = 70 ways for this case.

4. Since the two cases are mutually exclusive and cover every valid selection, add them (sum rule): 21 + 70 = 91.

Cross-check (complementary counting): the total number of ways to pick any 5 guests from all 9 friends is 9C5 = 126. Among these, the selections where both A and B attend together account for 7C3 = 35 (choosing the remaining 3 guests from the other 7 friends once A and B are both fixed). Removing these gives 126 − 35 = 91, confirming the case-based total.

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