Problem Solving

[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 usage of the term arrangements is misleading. The prompt seems intended to ask the following:

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.

The correct answer is B.

Mitch has a good point. We will follow his suggestion.

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

$$?\,\,\,:\,\,\,\# \,\,{\rm{pairs}}\,\,{\rm{among}}\,\,6\,\,{\rm{people}}$$
$${\rm{If}}\,\,{\rm{teams}}\,\,\left( {{\rm{pairs}}} \right)\,\,A,B,C\,\,{\rm{are}}\,\,{\rm{distinguishable:}}$$
$$\underbrace {C\left( {6,2} \right)}_{{\rm{team}}\,\,A}\,\,\,\, \cdot \,\,\,\underbrace {C\left( {4,2} \right)}_{{\rm{team}}\,\,B}\,\,\,\, = \,\,\,\,{{6 \cdot 5} \over 2} \cdot {{4 \cdot 3} \over 2} = 15 \cdot 6 = 90\,\,\,\,\,\left[ {{\rm{team}}\,\,{\rm{C}}\,\,{\rm{ = }}\,\,{\rm{remainder}}\,\,{\rm{players}}} \right]$$
$${\rm{Filtering}}\,\,\left( {{\rm{same}}\,\,{\rm{pairs}}\,\,{\rm{at}}\,\,{\rm{different}}\,\,{\rm{named}}\,\,{\rm{teams}}} \right)\,\,\,{\rm{:}}\,\,\,\,\,? = {{90} \over {3!}} = 15$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

=>
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