[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
In a sports club, 6 players are divided into 3 teams of 2 pl
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- Max@Math Revolution
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The usage of the term arrangements is misleading. The prompt seems intended to ask the following:
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.
The correct answer is B.
The first player selected can be paired with any of the 5 other players.How many ways can 6 players be divided into pairs?
A. 10
B. 15
C. 25
D. 30
E. 40
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.
The correct answer is B.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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- fskilnik@GMATH
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Mitch has a good point. We will follow his suggestion.
$${\rm{If}}\,\,{\rm{teams}}\,\,\left( {{\rm{pairs}}} \right)\,\,A,B,C\,\,{\rm{are}}\,\,{\rm{distinguishable:}}$$
$$\underbrace {C\left( {6,2} \right)}_{{\rm{team}}\,\,A}\,\,\,\, \cdot \,\,\,\underbrace {C\left( {4,2} \right)}_{{\rm{team}}\,\,B}\,\,\,\, = \,\,\,\,{{6 \cdot 5} \over 2} \cdot {{4 \cdot 3} \over 2} = 15 \cdot 6 = 90\,\,\,\,\,\left[ {{\rm{team}}\,\,{\rm{C}}\,\,{\rm{ = }}\,\,{\rm{remainder}}\,\,{\rm{players}}} \right]$$
$${\rm{Filtering}}\,\,\left( {{\rm{same}}\,\,{\rm{pairs}}\,\,{\rm{at}}\,\,{\rm{different}}\,\,{\rm{named}}\,\,{\rm{teams}}} \right)\,\,\,{\rm{:}}\,\,\,\,\,? = {{90} \over {3!}} = 15$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
$$?\,\,\,:\,\,\,\# \,\,{\rm{pairs}}\,\,{\rm{among}}\,\,6\,\,{\rm{people}}$$Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
In a sports club, 6 players are divided into pairs. How many such arrangements are possible?
A. 10
B. 15
C. 25
D. 30
E. 40
$${\rm{If}}\,\,{\rm{teams}}\,\,\left( {{\rm{pairs}}} \right)\,\,A,B,C\,\,{\rm{are}}\,\,{\rm{distinguishable:}}$$
$$\underbrace {C\left( {6,2} \right)}_{{\rm{team}}\,\,A}\,\,\,\, \cdot \,\,\,\underbrace {C\left( {4,2} \right)}_{{\rm{team}}\,\,B}\,\,\,\, = \,\,\,\,{{6 \cdot 5} \over 2} \cdot {{4 \cdot 3} \over 2} = 15 \cdot 6 = 90\,\,\,\,\,\left[ {{\rm{team}}\,\,{\rm{C}}\,\,{\rm{ = }}\,\,{\rm{remainder}}\,\,{\rm{players}}} \right]$$
$${\rm{Filtering}}\,\,\left( {{\rm{same}}\,\,{\rm{pairs}}\,\,{\rm{at}}\,\,{\rm{different}}\,\,{\rm{named}}\,\,{\rm{teams}}} \right)\,\,\,{\rm{:}}\,\,\,\,\,? = {{90} \over {3!}} = 15$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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- Max@Math Revolution
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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*6/6 = 15
Therefore, the answer is B.
Answer: B
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*6/6 = 15
Therefore, the answer is B.
Answer: B
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The number of ways to select the first team is 6C2 = (6 x 5)/2! = 15Max@Math Revolution 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 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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