Permutations and combinations help - confirm solution ?

[Eagerissac]

So I did the first 3 questions without any problems, but I'm stuck on 4.

I first drew what the arrangement might look like and got:

Left side:
T L I H I L I L

Right side:
I L I L T L H I

T = all together now
H = hot pink
I = IPA
L = lager

At first I thought I would do 2 * 1 * 3! * 3! because there are 2 options of beer and since the hot pink bottle cannot be adjacent to the all together bottle, it forces hot pink into position four, leaving only that 1 option. The 3! and 3! came from the three positions leftover for each beer type. I thought I would do 2 * 1 * 3! * 3! + 2 * 1 * 3! * 3! because there are two different sides I have to account for but I feel like I'm accidentally counting some positions too many times.

I'm not sure if the 2nd digit should be a 1 either. I would really appreciate some help or if someone told me what's wrong with my solution as I'm very confused as to how I would solve 4). I think I would've been fine if the question only asked for a specific side, but with both the left and right side I'm not sure what to do.

 

[Romsek]

I think the thing to do is get the arrangements of AT and HP sorted first

You want AT in only the first 4
AT L I L
I L AT L
L AT L I
L I L AT

The two that end in L can then have
I HP I L, or
I L I HP as the last 4

The one that ends in I can have
HP I L I, or
L I HP I as the last 4

The one that ends in AT can have
L I HP I as the last 4

so that's [MATH]2\cdot 2 + 2 + 1 = 7[/MATH] ways of arranging AT and HP within the 8 bottles.

Now each one of these arrangements has [MATH](3!)^2=36[/MATH] variations due to the other 6 bottles.

so we have a total of [MATH]7 \cdot 36 = 252[/MATH] total arrangements meeting #4's requirements.

 

[Romsek]

Maybe a less brute force way of doing is to first pick the slot for AT.

There are 4 possible slots. We then pick a slot for HP.
In 3 of the arrangements of AT we have 2 slots HP can appear in.
In 1 of them there is only one slot for HP to appear in.

Thus again we have 7 arrangements of AT and HP, leading to the same answer as in post #2