A firm is divided into four departments, each of which contains four people. If a project is to be assigned to a team of three people, none of which can be from the same department, what is the greatest number of distinct teams to which the project could be assigned?
A 4^3
B. 4^4
C. 4^5
D. 6(4^4)
E. 4(3^6)
OA:B
What is the flaw in this approach:
There are three people to be selected from 4 deptt. each employing 4 persons.
For the first place:
# of ways of choosing 1 deptt. out of 4=4C1 ways
# of ways of choosing 1 person from that deptt.=4P1 ways
So, # of ways of choosing a person for the first place: 4C1*4P1 ways.
Similarly,
# of ways of choosing for the second place: 3C1*4P1
# of ways of choosing for the third place: 2C1*4P1
Combining all the three steps,
Total number of ways of selecting a team of three people, none of which can be from the same department: 4C1*4P1*3C1*4P1*2C1*4P1= 6(4^4)
Thanks!
A 4^3
B. 4^4
C. 4^5
D. 6(4^4)
E. 4(3^6)
OA:B
What is the flaw in this approach:
There are three people to be selected from 4 deptt. each employing 4 persons.
For the first place:
# of ways of choosing 1 deptt. out of 4=4C1 ways
# of ways of choosing 1 person from that deptt.=4P1 ways
So, # of ways of choosing a person for the first place: 4C1*4P1 ways.
Similarly,
# of ways of choosing for the second place: 3C1*4P1
# of ways of choosing for the third place: 2C1*4P1
Combining all the three steps,
Total number of ways of selecting a team of three people, none of which can be from the same department: 4C1*4P1*3C1*4P1*2C1*4P1= 6(4^4)
Thanks!
