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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
Joined: 16 Aug 2015
Posts: 8017
GMAT 1: 760 Q51 V42
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22 Aug 2018, 01:45
00:00

Difficulty:

5% (low)

Question Stats:

84% (00:49) correct 16% (00:37) wrong based on 66 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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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" GMAT Club Legend Joined: 12 Sep 2015 Posts: 4009 Location: Canada Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb [#permalink] ### Show Tags 22 Aug 2018, 05:39 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 RELATED VIDEO FROM OUR COURSE _________________ Test confidently with gmatprepnow.com Director Joined: 20 Feb 2015 Posts: 762 Concentration: Strategy, General Management Re: Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb [#permalink] ### Show Tags 22 Aug 2018, 10:15 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 GMAT Club Legend Joined: 12 Sep 2015 Posts: 4009 Location: Canada Re: Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb [#permalink] ### Show Tags 22 Aug 2018, 10:20 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 _________________ Test confidently with gmatprepnow.com Intern Joined: 14 Sep 2014 Posts: 14 Location: India Schools: IIMA PGPX "21 Re: Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb [#permalink] ### Show Tags 22 Aug 2018, 10:45 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 Posted from my mobile device Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8017 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb [#permalink] ### Show Tags 24 Aug 2018, 00:52 => 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb  [#permalink]

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26 Aug 2018, 18:37
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.

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Re: Alice, Bob, Cindy, Dave and Eddie joined a three-person-a-side basketb   [#permalink] 26 Aug 2018, 18:37
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