Question about combinations using the "or"

Question

I want to form two groups that can each contain either 5 or 6 people. I have 9 female and 8 male volunteers that would like to participate in this formation. In how many ways can this be done?

This question might be really simple but I don't really understand how to deal with the "or". For this specific question the female and male count doesn't really matter but it does for the next few. But for now, I just want to understand the basics first.

For a single group of 5, it's easy for me to say that I have to calculate 17C5, but I'm not sure how I should think if I'm dealing with two groups that can have different amounts of people each.

Thanks

Answer 1

In general, or refers to a union of two events. So the inclusion-exclusion principle https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle tells us that we can count the number of elements in $A \cup B$, an or situation, by using $|A \text{ or } B| = |A| + |B| - |A \text{ and } B|$ where $| \cdot |$ denotes size.

Edit: In some cases, the intersection, corresponding to and, is empty. For example, I can't have a group that consists of five people and of six people, simultaneously. So in those cases, I can simply add the counts.

Added example: Let's say we want to count the number of pairs of (ordered) groups that can be made from a collection of 17 distinguishable marbles where each group consists of five or six marbles. Our possibilities are as follows:

• group A has five and group B has five
• group A has six and group B has five
• group A has five and group B has six
• group A has six and group B has six

For each of these, we can count the number of ways to make them.

• group A has five and group B has five: $C(17,5)\cdot C(12,5)$
• group A has six and group B has five: $C(17,6)\cdot C(11,5)$
• group A has five and group B has six: $C(17,5)\cdot C(12,6)$
• group A has six and group B has six: $C(17,6)\cdot C(11,6)$

Then the total number of possible groups (where the two groups are ordered) with this format should be the sum of each of the above since the scenarios are mutually exclusive. If we don't want to distinguish between the two groups, the two symmetric (five-six and six-five) scenarios "collapse" into one.