A firm is divided into four deptt.(P&C)

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A firm is divided into four deptt.(P&C)

by parveen110 » Fri Mar 21, 2014 5:05 am
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!
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by GMATGuruNY » Fri Mar 21, 2014 6:42 am
parveen110 wrote: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
Number of options for the first team member = 16. (Any of the 16 people.)
Number of options for the second team member = 12. (Any of the 12 people in a different department from the first team member's.)
Number of options for the third team member = 8. (Any of the 8 people in a different department from the first two members' departments.)
To combine these options, we multiply:
16*12*8.
Since the ORDER of the selections doesn't matter -- ABC is the same team as BCA -- we divide by the number of ways to ARRANGE the 3 members selected (3!):
(16*12*8)/(3*2*1) = 16*16 = 4�.

The correct answer is B.
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