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

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 -

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.

\(\left. \matrix{\\
{\rm{Full}}\,\,\,{\rm{:}}\,\,\,\,{\rm{2}}\,\,{\rm{students}}\,\,\left( {A,B} \right) \hfill \cr \\
{\rm{Swee}}\,\,\,{\rm{:}}\,\,\,{\rm{2}}\,\,{\rm{students}}\,\,\left( {C,D} \right) \hfill \cr \\
{\rm{Goal}}\,\,\,{\rm{:}}\,\,\,{\rm{3}}\,\,{\rm{students}}\,\,\left( {E,F,G} \right)\,\, \hfill \cr} \right\}\,\,\,\,\,{\rm{with}}\,\,{\rm{restrictions}}\,\,\,\left\{ \matrix{\\
\,D\,\,\, \Rightarrow \,\,\,B\,\,\,\left( * \right) \hfill \cr \\
\,C\,\,\, \Rightarrow \,\,\,E\,\,\,\left( {**} \right) \hfill \cr} \right.\)


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


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


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

Regards,
Fabio.

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

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

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.

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)