To solve this problem, let’s denote the friends as A, B, C, D, E, F, G, and H, where F and G are the feuding friends. We can approach this by first calculating the total number of ways to choose 5 friends from 8 without any restrictions and then subtract the cases where both F and G are selected.
Step 1: Calculate total combinations without restrictions
The number of ways to choose 5 friends out of 8 is given by the combination formula:
C(n, k) = n! / [k!(n – k)!]
For this scenario, it is C(8, 5), which simplifies to:
C(8, 5) = 8! / (5! * (8 – 5)!) = 56.
Step 2: Calculate combinations where both F and G are selected
If both F and G are invited, we only need to choose 3 more friends from the remaining 6 (A, B, C, D, E, H). The number of combinations would then be:
C(6, 3) = 6! / (3! * (6 – 3)!) = 20.
Step 3: Subtract the restricted cases from the total combinations
Finally, we need to subtract the cases where both F and G are selected from the total combinations. So, the total number of valid combinations is:
56 (total combinations) – 20 (both F and G) = 36.
Conclusion
Therefore, the total number of ways to invite 5 friends to the party, ensuring that F and G do not attend together, is 36.