I have two ways of doing this question but they give different answers. The reasoning in one of my methods must be flawed and I would appreciate any help in pointing it out.
Method 1. First assume balls are distinguishable then divide by 5! at the end.
Number of ways to arrange 5 balls with no constraint is $12\cdot11\cdot10\cdot9\cdot8 $
Number of ways to arrange 5 balls with at least 1 ball next to each other is $$ 10\cdot2\cdot10\cdot9\cdot8+2\cdot1\cdot10\cdot9\cdot8$$
The first part comes from putting the first ball in 1 of the 10 middle boxes then putting the second ball in one of the boxes next to it and putting the remaining 3 balls anywhere.
The second part comes from putting the first ball in one of the the two end boxes, then putting the second ball next to it and the remaining 3 anywhere.
and the final answer is
$$ \frac{12\cdot11\cdot10\cdot9\cdot8-(10\cdot2\cdot10\cdot9\cdot8+2\cdot1\cdot10\cdot9\cdot8)}{5!}$$
I strongly believe this method is flawed, can someone help me point out where I went wrong.
The other method which I assume to be correct is.
$$\binom{6}{5}+2\cdot\binom{6}{4}+\binom{6}{3}$$
This one comes from arranging the blue balls in between gaps.
You second calculation is correct. To do it the first way, you can use the principle of inclusion and exclusion, as I said. There are $12 \choose5$ ways to distribute $5$ balls. There are $11$ ways to place two balls next to one another, and then $10\choose3$ ways to distribute the remaining balls giving $${12 \choose5}-11\cdot{10\choose3}$$ ways so far. However, we have subtracted any distribution with two pairs of consecutive balls twice, so we have to add those back in. This is a bit tricky, since we could either have two pairs that don't touch, or a run of three. Once you've figured that out, you need to subtract out the distributions with three pairs. This can either be a pair and run of three or a run of four. Finally, you need to add back the distributions with four pairs, which can only be a run of five.