How does this work, if y cannot go in a "o" or "g" window how do you handle y?

Here's the setup:

G, O, P, R, Y

Rule 1: (G + P) x1 *exactly*
Rule 2: R x2 *exactly*
Rule 3: Y --> ~G, ~O
Rule 4: ~P --> O

CPos 3: (O or G) --> ~Y
CPos 4: ~O --> P

Rule 3 + CPos 4: Y --> ~O --> P

___ ___ ___ ___
P/O
___ ___ ___ ___
P/O
___ ___ ___ ___
P/O



Now for the question:

If O > G, then we could have:
O x3, G x2
-OR-
O x3, G x1
-OR-
O x2, G x1

But we know that Y cannot be in the same window as G or O (Rule 3).
So one window must have Y, and we must have O x2, and G x1.
And the window with G must have both G + O.

Let's update our Model:
O G ___ ___

O __ ___ ___

P __ ___ ___


Alright. Let's look at the Rules. Rule 1 says we must have G + P once. And we know that G occurs only once. So P must be in the same window as our only G.

And we'll have to stick Y in the window with P.


Let's update our Model again:
O G P ___

O ___ ___ ___

P Y ___ ___


Note that each window must have at least two colors.
This means that either an R or a P must appear in the window that has only an O (middle window).
At this point, we don't know for sure where the two R's go.

Update:
O G P R?

O R? P? ___

P Y R? ___


Now, on to the elimination:

Q 11:
(A) This is it. If we have place our R's in the top and bottom windows, then we can have only O and P in the middle window.
(B) Not possible. The window with G has O, which means it can't have P + R as well. (That'd be 4 colors.)
(C) Nope. Not possible to ONLY have G + P (remember, we need to place two O's).
(D) Nope. With G must come P.
(E) No. The window with G must have P as well.