To solve the problem of how many ways you can invite 5 friends to a party from a selection of 8 friends, especially with the stipulation that 2 particular friends (let’s call them A and B) cannot attend together, we can use combinatorial methods.
First, we should calculate the total combinations without any restrictions. The number of ways to choose 5 friends from 8 is given by the combination formula:
C(n, k) = n! / (k!(n-k)!)
where n is the total number of friends (8 in this case) and k is the number of friends to choose (5 here).
So, calculating it:
- C(8, 5) = 8! / (5!(8-5)!) = 8! / (5!3!)
- Calculating the factorials, we find:
- C(8, 5) = (8 × 7 × 6) / (3 × 2 × 1) = 56
This means without any restrictions, there are 56 ways to choose 5 friends from 8.
Next, we need to account for the restriction that friends A and B cannot both be invited together. To do this, we will calculate the number of combinations where A and B are both invited together and then subtract that from the total combinations.
If friends A and B are both attending, we have already invited 2 friends, which means we need to choose 3 more friends from the remaining 6 (since 2 out of the 8 have been already chosen). Thus, the number of ways to choose 3 friends from the remaining 6 is:
- C(6, 3) = 6! / (3!(6-3)!) = 6! / (3!3!)
- C(6, 3) = (6 × 5 × 4) / (3 × 2 × 1) = 20
So, there are 20 combinations where friends A and B are both invited.
Finally, we subtract this from the original total:
- Total valid combinations = Total combinations – Invalid combinations
- Total valid combinations = 56 – 20 = 36
Therefore, the total number of ways to invite 5 friends to the party, ensuring that friends A and B do not attend together, is 36.