eltonli wrote:

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?

A. 20

B. 30

C. 50

D. 56

E. 336

We can use the following formula:

Total number of ways to select the committee = # of ways with both inexperienced engineers selected + # of ways with two inexperienced engineers NOT selected

Thus:

# of ways with two inexperienced engineers NOT selected = Total number of ways to select the committee - # of ways with both inexperienced engineers selected

Total number of ways to select the committee:

8C3 = (8 x 7 x 6)/3! = 56 ways

Now let’s calculate the total number of ways to select the committee such that the two inexperienced engineers are both selected. One such occurrence would be:

NNE (N = inexperienced engineer and E = experienced engineer)

Since both inexperienced engineers have been selected, there is only 1 position left, and there are 6 experienced engineers to fill it. Thus, we have:

6C1 = 6

Thus, the number of ways to select the committee members with both inexperienced engineers NOT selected = 56 - 6 = 50.

Answer: C

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