[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
Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-si
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- Max@Math Revolution
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A
B
C
D
E
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The order in which we select the 3 starters does not matter.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
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
Cheers,
Brent
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When the answer choices are relatively small (as they are here), we should also consider just listing the possible outcomes.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
They are:
ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE
Answer: E
Cheers,
Brent
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!
They are:
ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE
Hence, the correct answer is E. Regards!
- Max@Math Revolution
- Elite Legendary Member
- Posts: 3991
- Joined: Fri Jul 24, 2015 2:28 am
- Location: Las Vegas, USA
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=>
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
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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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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