rc1979 wrote:

Ben needs to form a committee of 3 from a group of 8 engineers. If two of the engineers cannot serve together, how many different committees can Ben form?

Assume A and B are these two gentlemen in question.

Total ways of selecting engineers = C(8,3) = 56.

This includes

1. A included.

2. B included.

3. A and B included.

4. Nither A or B included.

Of course we need to discard the case of A and B included which is

C(6, 1) = 6.

Therefore, remaining cases, when these two gentlemen don't serve together = 56 - 6 = 50.

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