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Seven students are trying out for the school soccer team, on which the

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Seven students are trying out for the school soccer team, on which the  [#permalink]

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12 Jan 2015, 22:19
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Difficulty:

95% (hard)

Question Stats:

51% (01:54) correct 49% (01:13) wrong based on 300 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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Posts: 7100
Re: Seven students are trying out for the school soccer team, on which the  [#permalink]

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23 Dec 2017, 07:28
1
PathFinder007 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

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
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Re: Seven students are trying out for the school soccer team, on which the  [#permalink]

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12 Jan 2015, 23:13
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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

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Joined: 30 Apr 2017
Posts: 64
Re: Seven students are trying out for the school soccer team, on which the  [#permalink]

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29 Oct 2017, 06:45
EMPOWERgmatRichC wrote:
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

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???
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Joined: 29 Jun 2017
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Seven students are trying out for the school soccer team, on which the  [#permalink]

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Updated on: 23 Dec 2017, 09:34
soodia wrote:
EMPOWERgmatRichC wrote:
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

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???

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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Originally posted by sahilvijay on 23 Dec 2017, 07:10.
Last edited by sahilvijay on 23 Dec 2017, 09:34, edited 1 time in total.
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Re: Seven students are trying out for the school soccer team, on which the  [#permalink]

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24 Sep 2018, 21:41
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

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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Re: Seven students are trying out for the school soccer team, on which the &nbs [#permalink] 24 Sep 2018, 21:41
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