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

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by Max@Math Revolution » Wed Aug 22, 2018 12:40 am
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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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by Brent@GMATPrepNow » Wed Aug 22, 2018 4:38 am
Max@Math Revolution 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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Brent
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by Brent@GMATPrepNow » Wed Aug 22, 2018 9:21 am
Max@Math Revolution 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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by swerve » Thu Aug 23, 2018 11:06 am
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

Hence, the correct answer is E. Regards!
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by Max@Math Revolution » Thu Aug 23, 2018 11:48 pm
=>

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 = (5*4)/(1*2) = 10.

Therefore, the answer is E.
Answer: E
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by Scott@TargetTestPrep » Sun Aug 26, 2018 5:37 pm
Max@Math Revolution 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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