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Imagine lining up all 6 people:
Total ways = 6!
This counts every possible ordering.

Each group has 2 people, and inside a group
(A, B) is the same as (B, A)
So for each group, divide by 2!
There are 3 groups=> divide by (2!)^3

The groups are identical:
  • (Group1, Group2, Group3)
  • (Group2, Group1, Group3)
These are the same arrangement.
So divide by 3!

Final Calculation=
6! ÷ [(2!)^3× 3!]
=15

Ans (B)
ExpertsGlobal5
In how many ways can a group of 6 people be divided into three groups, with each group consisting of 2 people?

A. 12
B. 15
C. 60
D. 90
E. 120


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ExpertsGlobal5
In how many ways can a group of 6 people be divided into three groups, with each group consisting of 2 people?

A. 12
B. 15
C. 60
D. 90
E. 120
B is the correct answer choice.

Video explanation:

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6c2*4c2*2c2/3!.

3! since one same selection can be duplicated 6 times.

Hope it helps.

Thanks!
ExpertsGlobal5
In how many ways can a group of 6 people be divided into three groups, with each group consisting of 2 people?

A. 12
B. 15
C. 60
D. 90
E. 120


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I calculated as follows 2C6 * 2C4 * 2C2 = 90
How can I be sure that 6 should be divided ?
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Renchinjav
I calculated as follows 2C6 * 2C4 * 2C2 = 90
How can I be sure that 6 should be divided ?

Because the three groups are not labeled. For example, the same division AB, CD, EF is counted again as

AB, EF, CD
CD, AB, EF
CD, EF, AB
EF, AB, CD
EF, CD, AB

So every division is counted 3! = 6 times. Therefore, we divide 90 by 6.
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