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Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb

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Math Revolution GMAT Instructor
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Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb  [#permalink]

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New post 22 Aug 2018, 00:45
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

85% (00:37) correct 15% (00:29) wrong based on 59 sessions

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[Math Revolution GMAT math practice question]

Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketball tournament. In how many ways can be the three starters be chosen?

A. 5
B. 6
C. 8
D. 9
E. 10

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Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb  [#permalink]

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New post 22 Aug 2018, 04:39
Top Contributor
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketball tournament. In how many ways can be the three starters be chosen?

A. 5
B. 6
C. 8
D. 9
E. 10


The order in which we select the 3 starters does not matter.
For example, selecting Alice then Cindy and then Dave to be the starters is the SAME as selecting Cindy then Dave and then Alice to be the starters.
Since order does not matter, we can use COMBINATIONS

We can select 3 players from 5 players in 5C3 ways
5C3 = (5)(4)(3)/(3)(2)(1) = 10

Answer: E

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Re: Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb  [#permalink]

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New post 22 Aug 2018, 09:15
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketball tournament. In how many ways can be the three starters be chosen?

A. 5
B. 6
C. 8
D. 9
E. 10



Out of 5 , we need 3
which is
5c3 = \(\frac{5*4}{2}\) = 10
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Re: Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb  [#permalink]

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New post 22 Aug 2018, 09:20
Top Contributor
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketball tournament. In how many ways can be the three starters be chosen?

A. 5
B. 6
C. 8
D. 9
E. 10


When the answer choices are relatively small (as they are here), we should also consider just listing the possible outcomes.

They are:
ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE

Answer: E

Cheers,
Brent
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Re: Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb  [#permalink]

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New post 22 Aug 2018, 09:45
Number of ways choosing first person = 5 second person = 4 and number of ways of choosing third person = 3

Total ways = 5*4*3
Since order does not matter divide by 3!
= 10

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Re: Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb  [#permalink]

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New post 23 Aug 2018, 23:52
=>

The number of ways of choosing the three starters is the number of ways of choosing three people from a set of five people, where order does not matter and repetition is not allowed.
The number of possible selections of three starters is:
5C3 = 5C2 = \(\frac{(5*4)}{(1*2)}\) = \(10\).

Therefore, the answer is E.
Answer: E
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Re: Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb  [#permalink]

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New post 26 Aug 2018, 17:37
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketball tournament. In how many ways can be the three starters be chosen?

A. 5
B. 6
C. 8
D. 9
E. 10


The number of ways choosing 3 people from 5 (where order doesn’t matter) is 5C3 = (5 x 4 x 3)/(3 x 2) = 10.

Answer: E
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Re: Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb &nbs [#permalink] 26 Aug 2018, 17:37
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Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb

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