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04 Jan 2019, 01:18
Kudos
Bookmarks
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:
56% (01:42) correct
44%
(02:02)
wrong
based on 345
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[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: 277
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:
How many ways can 6 players be divided into pairs?
A. 10
B. 15
C. 25
D. 30
E. 40
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.
