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Janice has divided her department of 12 employees into...

by AAPL » Tue Mar 13, 2018 5:10 am
Janice has divided her department of 12 employees into 4 teams of 3 employees each to work on a new project. Four new employees are to be added to Janice's department, but the number of teams is to remain the same. If every employee must be on exactly 1 team, then the total number of different teams into which Janice divide the enlarged department is approximately how many times the number of different teams into which Janice could have divided the original department?

A. 16
B. 170
C. 340
D. 1820
E. 43,680

The OA is B.

I don't have clear this PS question. I appreciate if any expert explains it to me. Thank you so much.
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by regor60 » Tue Mar 13, 2018 6:48 am
AAPL wrote:Janice has divided her department of 12 employees into 4 teams of 3 employees each to work on a new project. Four new employees are to be added to Janice's department, but the number of teams is to remain the same. If every employee must be on exactly 1 team, then the total number of different teams into which Janice divide the enlarged department is approximately how many times the number of different teams into which Janice could have divided the original department?

A. 16
B. 170
C. 340
D. 1820
E. 43,680

The OA is B.

I don't have clear this PS question. I appreciate if any expert explains it to me. Thank you so much.
The original department comprises 12 employees to be distributed among 4 teams of 3.

The first team can select 3 members 12x11x10 ways. However, since position within the team doesn't matter, this needs to be divided by 3!. Likewise, the second team can select its 3 members from the remaining 9 9x8x7 ways, also divided by 3!.

Carrying this out, the calculation is 12!/(3!)^4

The new department now has 16 members to be distributed also among 4 teams, but now with 4 members each

The first team can be assembled 16x15x14x13 ways, now divide by 4! since order doesn't matter. The remaining 3 teams assembled as described above.

Carrying this out, the calculation is 16!/(4!)^4. Now, (4!)^4 = 4^4x(3!)^4, so can be rewritten 16!/(4^4)x(3!)^4

The question asks how many times this number is compared to the first number, so:

[16!/(4^4)x(3!)^4] / 12!/((3!)^4) = 16!/((12!)x(4^4)) = (16x15x14x13)/(2^8)

with some factoring out of 2's this reduces to (15x7x13)/8, which is approximately 170, B
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by GMATinsight » Tue Mar 13, 2018 6:50 am
AAPL wrote:Janice has divided her department of 12 employees into 4 teams of 3 employees each to work on a new project. Four new employees are to be added to Janice's department, but the number of teams is to remain the same. If every employee must be on exactly 1 team, then the total number of different teams into which Janice divide the enlarged department is approximately how many times the number of different teams into which Janice could have divided the original department?

A. 16
B. 170
C. 340
D. 1820
E. 43,680

The OA is B.

I don't have clear this PS question. I appreciate if any expert explains it to me. Thank you so much.
4 Teams of 3 employees each can be made in (12C3*9C3*6C3*3C3)/4!

4 Teams of 4 employees each can be made in (16C4*12C4*8C4*4C4)/4!

Now the factor to be calculated = (16C4*12C4*8C4*4C4)/4! / (12C3*9C3*6C3*3C3)/4! = 1820*495*70 / 220*84*20 = 170.625

Answer: option B
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