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financebro28
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Christy, Dora, Emily and Flora are playing a game that requires two te 16 Feb 2025, 01:27
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30% (00:57) correct 70% (01:00) wrong based on 260 sessions

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Christy, Dora, Emily and Flora are playing a game that requires two teams of two members each. In how many ways can they form the teams?

A. 2
B. 3
C. 6
D. 12
E. 24
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B
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Christy, Dora, Emily and Flora are playing a game that requires two te 16 Feb 2025, 13:06
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financebro28
Christy, Dora, Emily and Flora are playing a game that requires two teams of two members each. In how many ways can they form the teams?

A. 2
B. 3
C. 6
D. 12
E. 24
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Possibility 1: Christy & Dora vs Emily & Flora

Possibility 2: Christy & Emily vs Dora & Flora

Possibility 3: Christy & Flora vs Dora & Emily

Hence, there are three team possible.

Option C
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Christy, Dora, Emily and Flora are playing a game that requires two te 29 Mar 2025, 15:33
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What would be wrong with my approach as 4C2? Pls help
gmatophobia
financebro28
Christy, Dora, Emily and Flora are playing a game that requires two teams of two members each. In how many ways can they form the teams?

A. 2
B. 3
C. 6
D. 12
E. 24

Possibility 1: Christy & Dora vs Emily & Flora

Possibility 2: Christy & Emily vs Dora & Flora

Possibility 3: Christy & Flora vs Dora & Emily

Hence, there are three team possible.

Option C
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Christy, Dora, Emily and Flora are playing a game that requires two te 30 Mar 2025, 03:05
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SIndo12
What would be wrong with my approach as 4C2? Pls help
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The correct approach includes dividing by 2! => \(4C2 * 2C2 * \frac{1}{2!}\)

We divide by 2! (the factorial of the number of pairs) because the order of the teams does not matter. For example, if the 4 people are labeled 1, 2, 3, and 4, the team-selection (1,2); (3,4) would be considered the same as (3,4); (1,2), as there is no specific order assigned to the teams.


You can practice similar questions here: https://gmatclub.com/forum/distributing ... 56656.html
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Christy, Dora, Emily and Flora are playing a game that requires two te 18 Apr 2025, 02:12
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Krunaal when we use 'combinations', we do so when order doesn't matter, then why are we dividing the result by 2!?

Krunaal
SIndo12
What would be wrong with my approach as 4C2? Pls help


The correct approach includes dividing by 2! => \(4C2 * 2C2 * \frac{1}{2!}\)

We divide by 2! (the factorial of the number of pairs) because the order of the teams does not matter. For example, if the 4 people are labeled 1, 2, 3, and 4, the team-selection (1,2); (3,4) would be considered the same as (3,4); (1,2), as there is no specific order assigned to the teams.


You can practice similar questions here: https://gmatclub.com/forum/distributing ... 56656.html
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Christy, Dora, Emily and Flora are playing a game that requires two te 18 Apr 2025, 04:10
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kanikaa9
Krunaal when we use 'combinations', we do so when order doesn't matter, then why are we dividing the result by 2!?

Krunaal

The correct approach includes dividing by 2! => \(4C2 * 2C2 * \frac{1}{2!}\)

We divide by 2! (the factorial of the number of pairs) because the order of the teams does not matter. For example, if the 4 people are labeled 1, 2, 3, and 4, the team-selection (1,2); (3,4) would be considered the same as (3,4); (1,2), as there is no specific order assigned to the teams.


You can practice similar questions here: https://gmatclub.com/forum/distributing ... 56656.html
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While 4C2 and 2C2 are combinations (used to choose team members without caring about order within each team), multiplying them gives us arrangements of two teams - and that does involve order (Team A then Team B vs. Team B then Team A).

Since the teams are unlabeled and swapping them doesn’t give a new outcome, we divide by 2! to correct for double-counting.

Hope it helps.
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Christy, Dora, Emily and Flora are playing a game that requires two te 20 Apr 2025, 00:52
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Let me try to clear this out for you.
You are right, combinations formula is not taking into account orded INSIDE the team thus team 1 of (CD) is not beeing double counted with (DC).
Never the less, when you are selecting 2 from 4, you are considering all possibilities (not double counting inside a team) as follows:
AB
AC
AD
BC
BD
CD
how ever, this is not correct since you need a pair of teams, which means that if you select (AB) as team 1 even without ordering them inside the team, team 2 is already selectes as well (as it will be CD). thus, the first and last option in the list above is in reality the same case and it does not have to be double counted.
hope it helps!

kanikaa9
Krunaal when we use 'combinations', we do so when order doesn't matter, then why are we dividing the result by 2!?

Krunaal
SIndo12
What would be wrong with my approach as 4C2? Pls help


The correct approach includes dividing by 2! => \(4C2 * 2C2 * \frac{1}{2!}\)

We divide by 2! (the factorial of the number of pairs) because the order of the teams does not matter. For example, if the 4 people are labeled 1, 2, 3, and 4, the team-selection (1,2); (3,4) would be considered the same as (3,4); (1,2), as there is no specific order assigned to the teams.


You can practice similar questions here: https://gmatclub.com/forum/distributing ... 56656.html
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