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#45688
Setup and Rule Diagram Explanation

This is a Grouping Game: Defined-Fixed, Unbalanced: Overloaded, Numerical Distribution.

Aside from the first rule that serves to better establish the distribution, all of the rules are conditional. In games that rely solely on conditional rules, you are generally not able to make definitive inferences about where variables must be placed, but rather you must understand the relationships between certain variables and how they may or may not be grouped with one another. That means that a traditional diagram where definitive inferences are represented on the base is not going to occur. Instead, we must set the base of five slots, diagram the rules and make conditional inferences, and then proceed to attack the questions based on that information.

S06_Game_#2_setup_diagram 1a.png
S06_Game_#2_setup_diagram 1b.png
With exactly one student selected, there are only two possible parent-teacher distributions, namely fixed 2-2 and 1-3 distributions (3-1 is impossible because F and G cannot be selected together, and 0-4 is impossible because U and W cannot be selected together). Thus, there are only two possible distributions in the game:
S06_Game_#2_setup_diagram 2.png
However, these distributions only play a role in the last two questions.
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 akocher
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#102070
Can you elaborate on this please.

With exactly one student selected, there are only two possible parent-teacher distributions, namely fixed 2-2 and 1-3 distributions (3-1 is impossible because F and G cannot be selected together, and 0-4 is impossible because U and W cannot be selected together). Thus, there are only two possible distributions in the game:

I think I am getting more parent/teacher combos.
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 Dave Killoran
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#102072
Hi akocher,

Thanks for the question! So, to figure these out, let's look at all the info we have about the numbers in the game:

  • * The committee has exactly 5 members.

    * There are 3 parents, 3 students, and 4 teachers available for the 5 person committee. In other words, it's 12 into 5 in a broad sense.

    * The committee includes exactly 1 student. This is super helpful because it instantly reduces the number of options, and carves out exactly one of the spaces for a student every time. It also turns the game into a 7 into 4 exercise (3 parents + 4 teachers for the 4 remaining spaces on the committee).


So, this info takes us through the first rule of the game. Without considering the other rules yet, here is what our options for the 4 remaining spaces look like in terms of just filling the 4 parent/teacher spaces with any numbers (this first table will be just be the number possibilities, some will be impossible but I'll address that in the next table):

  • Parent ..... ..... ..... Teacher

    ..... 4 ..... ..... ..... ..... 0
    ..... 3 ..... ..... ..... ..... 1
    ..... 2 ..... ..... ..... ..... 2
    ..... 1 ..... ..... ..... ..... 3
    ..... 0 ..... ..... ..... ..... 4

Of course, the actual parameters of the game and the rules eliminate several of these outcomes:


  • Parent ..... ..... ..... Teacher

    ..... 4 ..... ..... ..... ..... 0 ..... ..... This one is out since there are only 3 parents!
    ..... 3 ..... ..... ..... ..... 1 ..... ..... This one is out since F and H can't be together, so 3 parents is impossible.
    ..... 2 ..... ..... ..... ..... 2
    ..... 1 ..... ..... ..... ..... 3
    ..... 0 ..... ..... ..... ..... 4 ..... ..... This one is out since U and W can't be together, so 4 teachers is impossible.

What that leaves us with is the "fixed 2-2 and 1-3 distributions" I reference in the explanation.

Please let me know if that helps make this clearer. Thanks!

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