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

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.

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

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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