Ben needs to form a committee of 3 from a group of 8 engineers. If two of the engineers are too inexperienced to serve together on the committee, how many different committees can Ben form?
Since 8 choose 3 is 56, we know the answer has to be A, B, or C. That narrows down our choices a lot already.
Now, imagine our 8 engineers look like this:| | | | | | | |
where the first two are the ones who are too inexperienced to work together. If we have these two together in a group, then we have 6 remaining options for the last member of the group. Thus, there are only 6 possible committees with these two together.
56 total possibilities - 6 groups with these two together = 50 committees.