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In a sports club, 6 players are divided into 3 teams of 2 players. How

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Math Revolution GMAT Instructor
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In a sports club, 6 players are divided into 3 teams of 2 players. How  [#permalink]

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New post 04 Jan 2019, 00:18
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

43% (01:12) correct 57% (01:32) wrong based on 49 sessions

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[Math Revolution GMAT math practice question]

In a sports club, 6 players are divided into 3 teams of 2 players. How many such arrangements are possible?

A. 10
B. 15
C. 25
D. 30
E. 40

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Re: In a sports club, 6 players are divided into 3 teams of 2 players. How  [#permalink]

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New post 04 Jan 2019, 04:24
The usage of the term arrangements is misleading. The prompt seems intended to ask the following:

Quote:
How many ways can 6 players be divided into pairs?

A. 10
B. 15
C. 25
D. 30
E. 40


The first player selected can be paired with any of the 5 other players.
Thus, the number of options for the first player selected = 5.
At this point, 4 players remain.
The next player selected can be paired with any of the 3 other players.
Thus, the number of options for the next player selected = 3.
The 2 remaining players must serve as the final pair.
To combine the options in blue, we multiply:
5*3 = 15.


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Re: In a sports club, 6 players are divided into 3 teams of 2 players. How  [#permalink]

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New post 04 Jan 2019, 05:11
1
The no. of ways of selecting 2 players from 6 is \(^6C_2\) ways.
The no. of ways of selecting 2 players from 4 is \(^4C_2\) ways.
The no. of ways of selecting 2 players from 2 is \(^2C_2\) ways.

Total no. of ways are \(\binom{6}{2} \binom{4}{2} \binom{2}{2}\)

15*6*1 = 90 ways. As the order doesn't matter here, we have to divide by 3! here.

\(\frac{90}{3!}\)

15 ways.

OPTION: B
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Re: In a sports club, 6 players are divided into 3 teams of 2 players. How  [#permalink]

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New post 06 Jan 2019, 17:25
=>
Since we need to choose \(2\) players out of \(6\) players for the first team, choose \(2\) players out of \(4\) players for the second team and choose \(2\) players out of \(2\) players for the last team, the number of possible choices, in this order is 6C2*4C2*2C2.
Since the order in which the three teams is chosen does not matter, we need to divide this by the number of arrangements of these \(3\) teams, which is \(3!\).
Thus, the number of possible arrangements is 6C2*4C2*2C2/3! = \(15*\frac{6}{6} = 15\)

Therefore, the answer is B.
Answer: B
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Re: In a sports club, 6 players are divided into 3 teams of 2 players. How &nbs [#permalink] 06 Jan 2019, 17:25
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