A plant manager must assign 10 new workers to one of five shifts. She

Question

A plant manager must assign 10 new workers to one of five shifts. She needs a first, second, and third shift, and two alternate shifts. Each of the shifts will receive 2 new workers. How many different ways can she assign the new workers?

A. 2430
B. 2700
C. 3300
D. 4860
E. 5400

A. 2,430
B. 2,700
C. 3,300
D. 4,860
E. 56,700

KarishmaB · Accepted Answer

Select 2 people for the first shift in 10C2 ways, for second shift in 8C2 ways, for third shift in 6C2 ways. 
You are left with 4 people for 2 alternate shifts. You need to split them into 2 groups which are identical. 
So select 2 people for one alternate shift in 4C2 ways and divide by 2! to un-arrange (the two alternate shifts are identical).

You get 10C2 * 8C2 * 6C2 * 4C2 / 2! = 56700

TestPrepUnlimited · Answer

If we are asking how many ways to assign all 10 of the workers, then it seems like none of the choices are correct. If we start from the first shift to the last, we have 10C2 options for the first shift, 8C2 options for the 2nd shift since the first shift took away 2 workers, and 6C2 for the 3rd shift etc. The two alternative shifts mean AB with CD is the same as CD with AB, so we split the last 4 people into two groups of 2 but divide by 2 to account for the alternation.

So the answer should be 10C2 * 8C2 * 6C2 * 4C2 * 2C2 / 2 = 56700.

ogabry · Answer

Can you explain? I also got a huge number that is not within the answer range.

AnkithaSrinivas · Answer

the order does not matter - hence we would divide it by n!

ogabry · Answer

Can you explain your logic? I couldn't divide by n!

sting8 · Answer

ScottTargetTestPrep chetan2u Bunuel TestPrepUnlimited GMATWhizTeam Can anyone please help with the explanation for this problem? I am not able to even get close to the answer choices range

Found a solution here https://gmat.economist.com/gmat-advice/ ... gmat-quant , but can't understand the approach of taking only 10C2 and not going ahead with 8C2*6C2*4C2

bv8562 · Answer

TestPrepUnlimited 
Why 10C2 * 8C2 * 6C2 *  4C4  = 18900 and  WHY NOT  10C2 * 8C2 * 6C2 * 4C2 * 2C2 ? Also, could you please help me understand what does this statement " two alternate shifts " mean?

bv8562 · Answer

Bunuel  Could you please share the solution for this one?

IanStewart · Answer

The "solution" at that link isn't even close to being correct, and the right answer isn't among the answer choices. Karishma's solution earlier in this thread is perfect (as usual!), so if you're curious to know whether you solved correctly, you can just refer to her post.

ScottTargetTestPrep · Answer

Response:

The solution in the provided link is not correct.

The 10C2 * 8C2 * 6C2 * 4C2 approach is correct. The workers for the first, second and third shift can be chosen in 10C2, 8C2 and 6C2 ways, respectively. Once those three shifts are formed, we have four workers remaining and we should form two teams of two from among these four workers. We can choose one of the teams in 4C2 ways and the remaining two workers will form the other team. Thus, there are 10C2 * 8C2 * 6C2 * 4C2 = 56,700 ways to form the shifts.

ishan_521101 · Answer

To solve the problem of assigning 10 new workers to five shifts
(first, second, third, and two alternate shifts), with each shift
receiving exactly 2 workers, we need to use combinations and account
for the indistinguishability of the two alternate shifts.

Step-by-Step Solution:

Total Workers and Shifts:
There are 10 workers.
There are 5 shifts, each receiving 2 workers.

Choosing 2 Workers for Each Shift:
The number of ways to choose 2 workers out of 10 for the first shift
is (10 C 2).
After assigning 2 workers to the first shift, 8 workers remain. The
number of ways to choose 2 workers out of these 8 for the second shift
is (8 C 2).
After assigning 2 workers to the second shift, 6 workers remain. The
number of ways to choose 2 workers out of these 6 for the third shift
is (6 C 2).
After assigning 2 workers to the third shift, 4 workers remain. The
number of ways to choose 2 workers out of these 4 for the fourth shift
is (4 C 2).
Finally, the number of ways to choose 2 workers out of the remaining 2
for the fifth shift is (2 C 2).

Calculating the Combinations:
(10 C 2) × (8 C 2) × (6 C 2) × (4 C 2) × (2 C 2)

Simplifying the Combinations:
(10 C 2) = 10! / (2!(10-2)!) = 10 × 9 / 2 × 1 = 45
(8 C 2) = 8! / (2!(8-2)!) = 8 × 7 / 2 × 1 = 28
(6 C 2) = 6! / (2!(6-2)!) = 6 × 5 / 2 × 1 = 15
(4 C 2) = 4! / (2!(4-2)!) = 4 × 3 / 2 × 1 = 6
(2 C 2) = 2! / (2!(2-2)!) = 2 × 1 / 2 × 1 = 1

Multiplying the Results:
45 × 28 × 15 × 6 × 1 = 113400

Adjusting for Overcounting:
Since the two alternate shifts are indistinguishable, we need to
divide by the number of ways to arrange these 2 shifts, which is 2!:
113400 / 2 = 56700

Therefore, the number of different ways the plant manager can assign
the 10 new workers to the five shifts is: 56700

IshaanJ · Answer

Answer according to this is 113400 ?

Kinshook · Answer

A plant manager must assign 10 new workers to one of five shifts. She needs a first, second, and third shift, and two alternate shifts. Each of the shifts will receive 2 new workers.

How many different ways can she assign the new workers?

The number of ways to assign new workers = 10C2*8C2*6C2*4C2*2C2 = 10!/8!2! * 8!/2!6! * 6!/2!4! * 4!/2!2! * 1 /2 = 10!/(2!)^5 / 2= 56700

IMO E

A plant manager must assign 10 new workers to one of five shifts. She

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A plant manager must assign 10 new workers to one of five shifts. She

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