pushpitkc wrote:
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. What is the total of the different committees possible?
A. 60
B. 81
C. 255
D. 405
E. 455
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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