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azamaka
An employer has 6 applicants for a programming position and 4 applicants for a manager position. If the employer must hire 3 programmers and 2 managers, what is the total number of ways the employer can make the selection?

a) 1,490
b) 132
c) 120
d) 60
e) 23

The programmers can be selected in 6C3 = 6!/3![(6-3)!] = (6 x 5 x 4)/3! = (6 x 5 x 4)/(3 x 2) = 20 ways.

The managers can be selected in 4C2 = 4!/[2!(4-2)!] = (4 x 3)/2! = 6 ways.

Thus, the total number of ways to select the group is 20 x 6 = 120.

Answer: C
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Why do we do 20x6 and not 20 + 6?
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Why do we do 20x6 and not 20 + 6?
Hi mtoti1.

That's a great question.

We multiply rather than add because we aren't adding the two sets of possible groups together. We are making combinations of the groups in the two sets of possible groups.

To see why we multiply to arrive at the numbers of combinations of possible groups, consider the following.

We can choose combinations of possible groups in the following way.

    We first choose one of the 20 possible groups of programmers. Then we choose one of the 6 possible groups of managers.

So, for every group of programmers, we have 6 ways to choose a group of managers.

    If we choose group 1 of the 20 possible groups of programmers, we have 6 ways to choose managers. So, we have 6 ways to choose after we choose group 1 of the programmers.

    If we choose group 2 of the 20 different groups of programmers, we again have 6 ways to choose managers. So, we have 6 ways to choose after we choose group 2 of the programmers.

    If we choose group 3 of the programmers, we have again 6 different ways to choose managers.

We have 20 total ways to choose programmers, and, for each of those, there are 6 different ways to choose managers.

So, once we know that we have 20 ways to choose programmers and 6 ways to choose managers, we can calculate that we have 20 x 6 = 120 ways to choose a group of programmers and then a group of managers.
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