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

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

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04 Jan 2019, 01:18
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52% (01:33) correct 48% (01:57) wrong based on 94 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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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior Manager Joined: 04 Aug 2010 Posts: 477 Schools: Dartmouth College Re: In a sports club, 6 players are divided into 3 teams of 2 players. How [#permalink] ### Show Tags 04 Jan 2019, 05:24 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. _________________ GMAT and GRE Tutor Over 1800 followers GMATGuruNY@gmail.com New York, NY If you find one of my posts helpful, please take a moment to click on the "Kudos" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Senior Manager Joined: 13 Jan 2018 Posts: 342 Location: India Concentration: Operations, General Management GMAT 1: 580 Q47 V23 GMAT 2: 640 Q49 V27 GPA: 4 WE: Consulting (Consulting) Re: In a sports club, 6 players are divided into 3 teams of 2 players. How [#permalink] ### Show Tags 04 Jan 2019, 06:11 4 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 _________________ ____________________________ Regards, Chaitanya +1 Kudos if you like my explanation!!! Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8017 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: In a sports club, 6 players are divided into 3 teams of 2 players. How [#permalink] ### Show Tags 06 Jan 2019, 18:25 => 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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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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27 Jan 2019, 19:38
MathRevolution wrote:
[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

The number of ways to select the first team is 6C2 = (6 x 5)/2! = 15

The number of ways to select the second team is 4C2 = (4 x 3)/2 = 6

The number ways to select the last team is 2C2 = 1

Since order does not matter for the team selection, the number of ways to select the teams is:

(15 x 6 x 1)/3! = 15 ways

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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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31 Jan 2019, 06:20
MathRevolution wrote:
=>
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.

Can we just not do 6C2, since the question is asking in how many ways we can select 2players from 6players.
Re: In a sports club, 6 players are divided into 3 teams of 2 players. How   [#permalink] 31 Jan 2019, 06:20
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# In a sports club, 6 players are divided into 3 teams of 2 players. How

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