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PathFinder007
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Seven students are trying out for the school soccer team, on which the 12 Jan 2015, 23:19
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56% (02:48) correct 44% (02:41) wrong based on 526 sessions

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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
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B
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Seven students are trying out for the school soccer team, on which the 13 Jan 2015, 00:13
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Hi PathFinder007,

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:
Show Spoiler
B

GMAT assassins aren't born, they're made,
Rich
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Seven students are trying out for the school soccer team, on which the 23 Dec 2017, 08:28
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PathFinder007
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
Show more


three category..
1) Full back - \(F_1 , F_2\).... NO restrictions
2) sweeper - \(S_3, S_4\)...
\(S_3\) plays if \(G_5\) plays
\(S_4\) plays if \(F_2\) plays
so restrictions for BOTH
3) Goalie - \(G_5, G_6,G_7\)... No restrictions

so lets look at the sweeper as the restrictions are there
let \(S_3\)play so ANY of two F and \(G_5\) in G .... so \(1*2*1=2\) ways
let \(S_4\) play so \(F_2\) in F and ANY of three in G .... so \(1*1*3=3\) ways

Total = 2+3=5 ways
General Discussion
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Seven students are trying out for the school soccer team, on which the 29 Oct 2017, 07:45
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EMPOWERgmatRichC
Hi PathFinder007,

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:
Show Spoiler
B

GMAT assassins aren't born, they're made,
Rich
Show more



tough one!
is there any algebraic solution for this?
is it possible we see such a question n real Gmat???
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Seven students are trying out for the school soccer team, on which the Updated on: 23 Dec 2017, 10:34
Originally posted by sahilvijay on 23 Dec 2017, 08:10.
Last edited by sahilvijay on 23 Dec 2017, 10:34, edited 1 time in total.
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soodia
EMPOWERgmatRichC
Hi PathFinder007,

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:
Show Spoiler
B

GMAT assassins aren't born, they're made,
Rich



tough one!
is there any algebraic solution for this?
is it possible we see such a question n real Gmat???
Show more





Let me try ->
a1a2 for F
a3a4 for S
a4-7 for G
a1 a2 a3 a4 a5 a6 a7 be the candidates { a4 will be there when a2 is there and a 3 when a5 is there => LOCK a2 and a5 in mind or paper}
F ------S ------G
a1-----a3-----a5,a6,a7 => 3 ways
a2-----a4-----a6,a7 only => 2 ways

total 5 ways - SHORT and SWEET -
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Seven students are trying out for the school soccer team, on which the 24 Sep 2018, 22:41
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Let’s call the students A, B, C, D, E, F, and G. There are 2 options for fullback (A or B), 2 options for sweeper (C or D), and 3 options for goalie (E, F, or G). Normally that would give us 2 x 2 x 3 = 12 options.

However, we have a second set of conditions. D can only play if B is used. And C will only play if E will also play. Since there are only 12 options, it’s easier to remove those that include D without B and C without E:

There’s only 5 options left!

ACE
ACF
ACG
ADE
ADF
ADG
BCE
BCF
BCG
BDE
BDF
BDG

If you chose (A), you may have crossed off too many options as you were going through them.

If you chose (C), this is merely the sum of the original options, which is not how a combination is calculated.

If you chose (D), this is still far too many options to satisfy both requirements.

If you chose (E), this is the number of options had there not been additional requirements.
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Seven students are trying out for the school soccer team, on which the 17 Dec 2018, 04:43
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PathFinder007
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
Show more
\(\left. \matrix{\\
{\rm{Full}}\,\,\,{\rm{:}}\,\,\,\,{\rm{2}}\,\,{\rm{students}}\,\,\left( {A,B} \right) \hfill \cr \\
{\rm{Swee}}\,\,\,{\rm{:}}\,\,\,{\rm{2}}\,\,{\rm{students}}\,\,\left( {C,D} \right) \hfill \cr \\
{\rm{Goal}}\,\,\,{\rm{:}}\,\,\,{\rm{3}}\,\,{\rm{students}}\,\,\left( {E,F,G} \right)\,\, \hfill \cr} \right\}\,\,\,\,\,{\rm{with}}\,\,{\rm{restrictions}}\,\,\,\left\{ \matrix{\\
\,D\,\,\, \Rightarrow \,\,\,B\,\,\,\left( * \right) \hfill \cr \\
\,C\,\,\, \Rightarrow \,\,\,E\,\,\,\left( {**} \right) \hfill \cr} \right.\)


\(?\,\,\,:\,\,\,\# \,\,\left( {{\rm{Full}}\,,\,\,{\rm{Swee}}\,,\,{\rm{Goal}}} \right)\,\,{\rm{choices}}\)


\({\rm{?}}\,\,\,{\rm{:}}\,\,\,\underline {{\rm{organized}}}\,\,{\rm{manual}} \,\,{\rm{work}}\,\,{\rm{technique}}\,\,\,\,\left\{ \matrix{\\
\,\left( {A,C,E} \right)\,\,\,\left( {**} \right)\,\,\,1. \hfill \cr \\
\,\left( {A,D,{\rm{no!}}} \right)\,\,\,\left( * \right) \hfill \cr \\
\,\left( {B,C,E} \right)\,\,\,\left( {**} \right)\,\,\,2. \hfill \cr \\
\,\left( {B,D,E} \right)\,\,\,\,\left( * \right)\,\,\,\,3.\, \hfill \cr \\
\,\left( {B,D,F} \right)\,\,\,\,\left( * \right)\,\,\,\,4. \hfill \cr \\
\,\left( {B,D,G} \right)\,\,\,\,\left( * \right)\,\,\,\,5. \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,? = 5\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Seven students are trying out for the school soccer team, on which the 15 Dec 2019, 20:09
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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
Show more

If the fullback is the first student, then the sweeper must be the third student (since the fourth student won’t play unless the fullback is the second student). Since the sweeper is the third student, goalie must be the fifth student (since the third student won’t play unless goalie is the fifth student). Thus, we see that there is only one way to fill the positions if the fullback is the first student.

If the fullback is the second student, the sweeper can be the third or fourth students.

If the sweeper is the third student, the goalie must be the fifth student (same reason as above); therefore we see that there is only one way to fill the positions if the fullback is the second student and the sweeper is the third student.

If the sweeper is the fourth student (which is possible since the fullback is the second student), then the goalie can be any one of the remaining three students. Therefore, we see that there are three ways to fill the positions if the fullback is the second student and the sweeper is the fourth student.

In total, there are 1 + 1 + 3 = 5 ways to fill the positions.

Answer: B
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Seven students are trying out for the school soccer team, on which the 15 Dec 2019, 21:30
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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
Show more

3 positions and their respective try-outs:
F: 1, 2
S: 3, 4
G: 5, 6, 7

- Case choose 2,4:
F: 2
S: 4
G: 5 / 6/ /7 -> 3 options

- Case choose 3,5:
F: 1/2
S: 3
G: 5 -> 2 options
==> total 3 + 2 = 5 options
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Seven students are trying out for the school soccer team, on which the 30 Jan 2022, 08:08
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OE:
Let’s call the students A, B, C, D, E, F, and G. There are 2 options for fullback (A or B), 2 options for sweeper (C or D), and 3 options for goalie (E, F, or G). Normally that would give us 2 x 2 x 3 = 12 options.

However, we have a second set of conditions. D can only play if B is used. And C will only play if E will also play. Since there are only 12 options, it’s easier to remove those that include D without B and C without E:

There’s only 5 options left!

If you chose (A), you may have crossed off too many options as you were going through them.

If you chose (C), this is merely the sum of the original options, which is not how a combination is calculated.

If you chose (D), this is still far too many options to satisfy both requirements.

If you chose (E), this is the number of options had there not been additional requirements.
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Let, the players be 1,2,3,4,5,6,7.
1. If FB=1, then SW=3, and GL=5,6,7. Depending on GL, there are 3 ways.
2. If FB=2, then SW=4, and GL=6,7. Depending on GL, there are 2 ways. So, total 3+2=5 ways (B)
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