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04 Jan 2019, 01:18
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
55% (01:43) correct
45%
(01:59)
wrong
based on 388
sessions
History
Date
Time
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Not Attempted Yet
[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 players. How many such arrangements are possible?
A. 10
B. 15
C. 25
D. 30
E. 40
eswarchethu135
GMAT 2: 640 Q49 V27
Posts: 276
04 Jan 2019, 06:11
Kudos
Bookmarks
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
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
General Discussion
04 Jan 2019, 05:24
Kudos
Bookmarks
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.
Quote:
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.
