There are a couple of different ways to approach the 'math' in this question. Here's another way that breaks the possibilities into smaller groups:
We're told t0 form a committee of 3 from a total of 8 engineers. HOWEVER, two of the engineers are too inexperienced to be on the committee together, so we cannot allow that...
Let's call the two inexperienced engineers A and B; then there are the other 6 engineers who DO have enough experience.
There are now 3 options when it comes to forming a committee:
1) Engineer "A" and 2 experienced engineers.
2) Engineer "B" and 2 experienced engineers.
3) 3 experienced engineers
For the first option, we can choose 2 engineers from the 6 experienced engineers = 6C2 = 6!/(2!4!) = 15 options.
For the second option, we can choose 2 engineers from the 6 experienced engineers = 6C2 = 6!/(2!4!) = 15 options.
For the third option, we can choose 3 engineers from the 6 experienced engineers = 6C3 = 6!/(3!3!) = 10 options.
15+15+10 = 50 total options.
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