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.

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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