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

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MathRevolution
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In a sports club, 6 players are divided into 3 teams of 2 players. How [#permalink] New post   04 Jan 2019, 01:18
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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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B
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Re: In a sports club, 6 players are divided into 3 teams of 2 players. How [#permalink] New post   04 Jan 2019, 06:11
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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] New post   06 Jan 2019, 18:25
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=>
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 [#permalink] New post   21 Jan 2021, 19:30
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pk6969 wrote:
Brian123
but while selecting(combination formula) we anyway exclude the order. only while applying permutations order is there.


pk6969, the formulas that we have applied are for selecting players for each team. eg - for the first team the formular is 6c2, for the 2nd team the formula is 4c2 and so on.
Remember, the order of the teams is not important and if you were only selecting one team with the given conditions, 6C2 would have been enough, but as we are selecting 3 teams, we need to divide by 3! as the order in which the teams are selected/listed is not important. Consider the players A,B,C,D,E &F.

Team 1 - AB (Remember you picked this by doing 6C2)
Team 2 - CD (Picked this by doing 4c2)
Team 3 - EF (picked by doing 2c2)

Now it doesnt matter if you list say AB,CD and EF or CD, EF and AB. This is why we divide by 3!

The best approach is to understand the meaning behind the formula and not simply applying them! Hope this helps! :)
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Re: In a sports club, 6 players are divided into 3 teams of 2 players. How [#permalink] New post   04 Jan 2019, 05:24
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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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B
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Re: In a sports club, 6 players are divided into 3 teams of 2 players. How [#permalink] New post   27 Jan 2019, 19:38
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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

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


Can we just not do 6C2, since the question is asking in how many ways we can select 2players from 6players.
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Re: In a sports club, 6 players are divided into 3 teams of 2 players. How [#permalink] New post   22 Oct 2020, 05:42
Hi experts why are we dividing by 3!?can you please post links for understanding the same?
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Re: In a sports club, 6 players are divided into 3 teams of 2 players. How [#permalink] New post   20 Jan 2021, 13:39
MathRevolution why did we divide it by 3!?
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Re: In a sports club, 6 players are divided into 3 teams of 2 players. How [#permalink] New post   20 Jan 2021, 21:14
pk6969 wrote:
MathRevolution why did we divide it by 3!?


pk6969 Nistha86
, that is because the order in which the teams are listed does not matter. You can call your teams - A,B and C or B,C and A, its all the same.
Hope this leps!
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Re: In a sports club, 6 players are divided into 3 teams of 2 players. How [#permalink] New post   21 Jan 2021, 04:03
Brian123
but while selecting(combination formula) we anyway exclude the order. only while applying permutations order is there.
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Re: In a sports club, 6 players are divided into 3 teams of 2 players. How [#permalink]
21 Jan 2021, 04:03
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