Q22

[ohthatpatrick]

If C3 has G, we start with
1: __ __ +
2: __ __ +
3: G __ +

Did G have a rule?

Yes, it was that G cannot appear with P. Okay, so we know P is not in C3.

1: __ __ +
2: __ __ +
3: G __ + (~P)

So where is P?

We only get one P, remember. Could it go in either C1 or C2?

No, actually, because EVERYTHING in C2 also has a copy in C1.

If I say that C2 recycles paper, then C1 also recycles paper. I can't let that happen because I'm only allowed to have one center recycle paper.

So that means that P must be in C1 but NOT in C2.

1: P __ +
2: __ __ + (~P)
3: G __ + (~P)

So who CAN I put in C2?

I can't put G, or else it would copy a G into C1, and then we'd have the forbidden PG.

1: P __ +
2: __ __ + (~P, ~G)
3: G __ + (~P)

So we have W, N, and T.

I need to pick at least two of them. Actually, COULD I still pick all three?

Could W, N, and T all go into C2?

Nope, because then they'd have to all be copied into C1, giving C1 four things.

So I need to pick EXACTLY two out of W, N, and T to give to C2.
That means it would be
WT
WN
or
NT

But one of those is actually breaking a rule.

WT breaks rule 1. If you have wood, you must have newsprint. So that scenario is out.

That means it would be
WN or NT going in C2.

Either way you slice it, N is involved. So (B) is the answer.

Hope this helps.