An enterprise has five departments. There are three managers in each

I answered this type of question correctly for first time! (Go me).

The way I think about it is:
Number of options for first manager = 15
for second manager= 12
for third= 9
fourth= 6.
But 15*12*9*6 is permutation and as we don't care about order here since we are picking teams (can pick teams in any order, the teams will remain the same unless specified about order).

Therefore we divide by 4! as 4 managers can be arranged in 4! way. Answer will be =(15*12*9*6)/4! = 405.

15*12*9*6/5! = 81 —> B


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Since maximum of 1 manager can be chosen from each department. The question comes down to:
1. Selecting 4 dept out of 5
2. Select 1 manager out of 3 in each of the departments chosen in (1) above.

Ways to choose 4 departments = 5
Ways to choose 1 manager out fo 3 = 3

Since there are a total of 4 departments, the total possibilities = 5x3x3x3x3 = 405

\(C^4_5*3^4=405\):

5C4 is the number of ways to choose which 4 departments out of 5 will delegate managers to the committee. Each of those 4 chosen departments can send any of the three members it has, so 3*3*3*3 = 3^4.

Answer: D.
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Let's assume the managers are chosen in following ways from the different departments -

D1: 0 nos selected (1 way), and
D2: 1 nos selected (3C1 way), and
D3: 1 nos selected (3C1 way), and
D4: 1 nos selected (3C1 way), and
D5: 1 nos selected (3C1 way).

Or,

Same pattern is repeated in following manner -

D2: 0 nos selected (1 way), and
D3: 1 nos selected (3C1 way), and
D4: 1 nos selected (3C1 way), and
D5: 1 nos selected (3C1 way), and
D1: 1 nos selected (3C1 way).

And so on.

Therefore, total nos. in this combination will be -

= { 1 x 3C1 x 3C1 x 3C1 x 3C1 } x 5
= 81 x 5
= 405.

Number of ways to select four elements from 5(order doesn't matter here):
5C4 = 5
Number of combinations of 4 elements with 3 options in each slot:
3(3)(3)(3)=3⁴
Total = 5C4(3⁴)=5(81)= 405

Ans. D

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Given: An enterprise has five departments. There are three managers in each of the five departments. A committee of four has to be made such that not more than one manager comes from each department.
Asked: What is the total of the different committees possible?

Ways to chose 4 departments = 5C4 = 5
Ways to chose managers from the 4 departments (3 choices each) = 3^4 = 81
Ways to chose committees = 5*81 = 405

IMO D

Solution:

The number of total of the different committees can be formed is:

(3C1 x 3C1 x 3C1 x 3C1 x 3C0) x 5C4 = 3 x 3 x 3 x 3 x 1 x 5 = 405

(Note: The four factors of 3C1 mean four of the five departments each can choose 1 manager from the 3 managers they have to serve in the committee, while the factor 3C0 means the remaining department can’t choose any manager since the committee can only have 4 people. Since 4 of the 5 departments can send their managers to serve in the committee, the factor 5C4 is the number of ways to choose 4 departments from 5 available departments.)

Alternate Solution:

The first manager can be any one of the 3 x 5 = 15 managers. The second manager cannot be in the same department as the first manager; thus, there are 15 - 3 = 12 choices for the second manager. Following the same logic, there are 12 - 3 = 9 and 9 - 3 = 6 choices for the third and fourth managers, respectively. If the order were important, there would have been 15 x 12 x 9 x 6 ways to form the committee. However, the order is not important and thus, the actual number of ways to form the committee is (15 x 12 x 9 x 6) / 4! = (15 x 12 x 9 x 6) / (4 x 3 x 2 x 1) = 15 x 9 x 3 = 405.

Answer: D
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Step 1: choose which 4 departments will send a person to be on the committee out of the 5 total departments

“5 choose 4” = 5 different ways


And

Step 2: for each one of those 5 different ways, we end up with 4 different departments. Each department can send 1 person to be on the team.


Dept 1 - 3 available people

Dept 2 - 3 available people

Dept 3 - 3 available people.

Dept 4 - 3 available people

The number of different combinations we can have with 1 person from each of these departments is:

3 * 3 * 3 * 3 = 81


(5) * (81) = 405 ways

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