## ACM group organization

Hi there,

I have 4 groups (g1,g2,g3,g4) and I want to take from each group one question per examGroup.

That is:
One exam group (for my students, examGroup A) contains:
- first question from q1
- first quesiton from q2
- first question from q3
- first question from q4

Then, next group (for my students, examGroup B) contains:
- second question from q1
- second question from q2
- second question from q3
- second question from q4

and further.

Please, can you provide instructions how to realize this scenario?

Thank you,
M

### Replies (7)

#### RE: ACM group organization - Added by Frédéric Bréalover 4 years ago

with AMC 1.3.0 or AMC test, after reading the documentation, i would try this

first solution : the easiest ; 4 projects, no shuffle

• Create 4 projects A, B...
• Project A
 \insertgroupfrom[1]{g1}{0}
\insertgroupfrom[1]{g2}{0}

• Project B
 \insertgroupfrom[1]{g1}{1}
\insertgroupfrom[1]{g2}{1}


Documentation : \insertgroupfrom[n]{groupname}{i} The command does the same as \insertgroup[n]{groupname}, starting from element at index i (the first element has index 0).

second solution ; 4 projects , shuffle

• Create 4 projects A, B...
• Project A
\cleargroup{all}
\copygroupfrom[1]{g1}{all}{0}
\copygroupfrom[1]{g2}{all}{0}
\setgroupmode{all}{withreplacement} % if you comment -> no shuffle
\insertgroup{all}

• Project B
\cleargroup{all}
\copygroupfrom[1]{g1}{all}{1}
\copygroupfrom[1]{g2}{all}{1}
\setgroupmode{all}{withreplacement} % if you comment -> no shuffle
\insertgroup{all}


Documentation : \copygroupfrom[n]{groupA}{groupB}{i} The command does the same as \copygroup[n]{groupA}{groupB} starting from element at index i (the first element has index 0).

Third solution : 1 project ; \AMCStudentnumber, \ifthenelse and modulo
not easy to code for me but I would try out

#### RE: ACM group organization - Added by Frédéric Bréalover 4 years ago

may be easier with a latex counter, i'll attempt

#### RE: ACM group organization - Added by Alexis Bienvenüeover 4 years ago

You can also use something like the following code (where the \q QXX texts are to be replaced by full question environments).

\documentclass{article}
\usepackage{automultiplechoice}
\begin{document}

\setdefaultgroupmode{cyclic}

\def\q{\par question~}

\element{g1}{\q G1A}
\element{g1}{\q G1B}
\element{g1}{\q G1C}
\element{g1}{\q G1D}

\element{g2}{\q G2A}
\element{g2}{\q G2B}
\element{g2}{\q G2C}
\element{g2}{\q G2D}

\element{g3}{\q G3A}
\element{g3}{\q G3B}
\element{g3}{\q G3C}
\element{g3}{\q G3D}

\element{g4}{\q G4A}
\element{g4}{\q G4B}
\element{g4}{\q G4C}
\element{g4}{\q G4D}

\onecopy{10}{

{\bf\Large One copy}

\cleargroup{mixed}
\copygroup[1]{g1}{mixed}
\copygroup[1]{g2}{mixed}
\copygroup[1]{g3}{mixed}
\copygroup[1]{g4}{mixed}

Questions:
\shufflegroup{mixed}\insertgroup{mixed}
}

\end{document}


#### RE: ACM group organization - Added by Miralem Mehicover 4 years ago

Alexis Bienvenüe wrote:

You can also use something like the following code (where the q QXX texts are to be replaced by full question environments).
[...]

It provides the same questions. That is, the questions are repeated. I do not need shuffle, the questions are already ordered in their groups.

#### RE: ACM group organization - Added by Alexis Bienvenüeover 4 years ago

the questions are repeated.

What do you mean?

I do not need shuffle

Just remove \shufflegroup{mixed} then.

#### RE: ACM group organization - Added by Miralem Mehicover 4 years ago

Alexis Bienvenüe wrote:

the questions are repeated.

What do you mean?

I do not need shuffle

Just remove shufflegroup{mixed} then.

Well, it doesn't work:

\onecopy{4}{

\cleargroup{all}
\copygroupfrom1{g0}{all}
\copygroupfrom1{g1}{all}
\copygroupfrom1{g2}{all}
\copygroupfrom1{g3}{all}
%\setgroupmode{all}{withreplacement} % if you comment -> no shuffle
\insertgroup{all}

#### RE: ACM group organization - Added by Miralem Mehicover 4 years ago

The only option I found is to use:


\makeatletter
\newcommand{\serialnum}{\the\AMCid@etud}
\makeatother

\onecopy{4}{

\cleargroup{all}
\copygroupfrom[1]{g0}{all}{\serialnum}
\copygroupfrom[1]{g1}{all}{\serialnum}
\copygroupfrom[1]{g2}{all}{\serialnum}
\copygroupfrom[1]{g3}{all}{\serialnum}
%\setgroupmode{all}{withreplacement} % if you comment -> no shuffle
\insertgroup{all}



but then the questions are displayed with following order (#1,#2,#3,#0). If I would be able to rearrange the order to have (#0,#1,#2,#3) that will be perfect.