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



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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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

=>
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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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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Can we just not do 6C2, since the question is asking in how many ways we can select 2players from 6players.

Hi experts why are we dividing by 3!?can you please post links for understanding the same?