• NEW! FREE Beat The GMAT Quizzes
Hundreds of Questions Highly Detailed Reporting Expert Explanations
• 7 CATs FREE!
If you earn 100 Forum Points

Engage in the Beat The GMAT forums to earn
100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Seven students are trying out for the school soccer team, on tagged by: alanforde800Maximus ##### This topic has 3 expert replies and 0 member replies ## Seven students are trying out for the school soccer team, on ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult Seven students are trying out for the school soccer team, on which there are three available positions: fullback, sweeper, and goalie. Each student can only try out for one position. The first two students are trying out for fullback. The next two students are trying out for sweeper. The remaining three students are trying out for goalie. However, the fourth student will only play if the second student is also on the team, and the third student will only play if the fifth student is on the team. How many possible combinations of students are there to fill the available positions? A 3 B 5 C 7 D 10 E 12 ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15348 messages Followed by: 1864 members Upvotes: 13060 GMAT Score: 790 Top Reply alanforde800Maximus wrote: Seven students are trying out for the school soccer team, on which there are three available positions: fullback, sweeper, and goalie. Each student can only try out for one position. The first two students are trying out for fullback. The next two students are trying out for sweeper. The remaining three students are trying out for goalie. However, the fourth student will only play if the second student is also on the team, and the third student will only play if the fifth student is on the team. How many possible combinations of students are there to fill the available positions? A 3 B 5 C 7 D 10 E 12 To keep track of the number of options for each position, draw a TREE. Start with the MOST RESTRICTED position, which is SWEEPER. If Sweeper = 3, then Goalie = 5. If Sweeper = 4, then Fullback = 2. Here's the tree so far: Now complete the tree, drawing the number of options for the remaining position in each case: The number of ways to choose the players is equal to the number of boxed outcomes on the right. Total ways = 5. The correct answer is B. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 alanforde800Maximus wrote: Seven students are trying out for the school soccer team, on which there are three available positions: fullback, sweeper, and goalie. Each student can only try out for one position. The first two students are trying out for fullback. The next two students are trying out for sweeper. The remaining three students are trying out for goalie. However, the fourth student will only play if the second student is also on the team, and the third student will only play if the fifth student is on the team. How many possible combinations of students are there to fill the available positions? A 3 B 5 C 7 D 10 E 12 $$\left. \matrix{ {\rm{Full}}\,\,\,{\rm{:}}\,\,\,\,{\rm{2}}\,\,{\rm{students}}\,\,\left( {A,B} \right) \hfill \cr {\rm{Swee}}\,\,\,{\rm{:}}\,\,\,{\rm{2}}\,\,{\rm{students}}\,\,\left( {C,D} \right) \hfill \cr {\rm{Goal}}\,\,\,{\rm{:}}\,\,\,{\rm{3}}\,\,{\rm{students}}\,\,\left( {E,F,G} \right)\,\, \hfill \cr} \right\}\,\,\,\,\,{\rm{with}}\,\,{\rm{restrictions}}\,\,\,\left\{ \matrix{ \,D\,\,\, \Rightarrow \,\,\,B\,\,\,\left( * \right) \hfill \cr \,C\,\,\, \Rightarrow \,\,\,E\,\,\,\left( {**} \right) \hfill \cr} \right.$$ $$?\,\,\,:\,\,\,\# \,\,\left( {{\rm{Full}}\,,\,\,{\rm{Swee}}\,,\,{\rm{Goal}}} \right)\,\,{\rm{choices}}$$ $${\rm{?}}\,\,\,{\rm{:}}\,\,\,{\rm{manual}}\,\,\underline {{\rm{organized}}} \,\,{\rm{work}}\,\,{\rm{technique}}\,\,\,\,\left\{ \matrix{ \,\left( {A,C,E} \right)\,\,\,\left( {**} \right)\,\,\,1. \hfill \cr \,\left( {A,D,{\rm{no!}}} \right)\,\,\,\left( * \right) \hfill \cr \,\left( {B,C,E} \right)\,\,\,\left( {**} \right)\,\,\,2. \hfill \cr \,\left( {B,D,E} \right)\,\,\,\,\left( * \right)\,\,\,\,3.\, \hfill \cr \,\left( {B,D,F} \right)\,\,\,\,\left( * \right)\,\,\,\,4. \hfill \cr \,\left( {B,D,G} \right)\,\,\,\,\left( * \right)\,\,\,\,5. \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,? = 5$$ This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br ### GMAT/MBA Expert Elite Legendary Member Joined 23 Jun 2013 Posted: 10197 messages Followed by: 496 members Upvotes: 2867 GMAT Score: 800 Hi alanforde800Maximus, This question uses what's called "Formal Logic" (a concept that you would see repeatedly on the LSAT, but rarely on the GMAT). You can answer it with some drawings and careful note-taking. Based on the information in the prompt, we have seven players (A,B,C,D,E,F,G). We're asked for the total groups of 3 that can be formed with the following restrictions: 1) Only 1 player per position 2) A and B are trying out for fullback C and D are trying out for sweeper E, F and G are trying out for goalie 3) D will only play if B is also on the team C will only play if E is also on the team The big restrictions are in the two 'formal logic' rules.... -We can put E with ANYONE, but we can only put in C if E is ALSO there. -We can put B with ANYONE, but we can only put in D if B is ALSO there. By extension.... D can NEVER be with A (because then B would not be in the group) C can NEVER be with F or G (because then E would not be in the group) As such, there are only a few possibilities. We can have... ACE BCE BDE BDF BDG Final Answer: B GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0

Available with Beat the GMAT members only code

• Get 300+ Practice Questions

Available with Beat the GMAT members only code

• Award-winning private GMAT tutoring
Register now and save up to \$200

Available with Beat the GMAT members only code

### Top First Responders*

1 Ian Stewart 57 first replies
2 Brent@GMATPrepNow 31 first replies
3 Jay@ManhattanReview 29 first replies
4 GMATGuruNY 21 first replies
5 ceilidh.erickson 15 first replies
* Only counts replies to topics started in last 30 days
See More Top Beat The GMAT Members

### Most Active Experts

1 Scott@TargetTestPrep

Target Test Prep

199 posts
2 Max@Math Revolution

Math Revolution

84 posts
3 Brent@GMATPrepNow

GMAT Prep Now Teacher

69 posts
4 Ian Stewart

GMATiX Teacher

65 posts
5 GMATGuruNY

The Princeton Review Teacher

40 posts
See More Top Beat The GMAT Experts