Simple answer. As long as the final value is not less than 15A and you meet all the requirements of 240.4(B), then yes.
Complex answer. I don't know why I'm even addressing this question, but I'll play.
After this answer please allow us to continue discussing 110.14(C) and 310.15(B), because that's what were trying to resolve here. I can see how this could be considered relevant, but it's mostly just a distraction as far as I'm concerned.
For one, I assume you are NOT paralleling four 12-2's for a single 20A circuit, because that would be certifiable, lol.
I assume you mean
four 12-2 NM cables for
four 20A circuits.
Bundling them through a fire stopped hole would NOT require you to derate (adjust) in the manner of having more than three CCC's.
It would require you to derate (correct) in the manner of ambient temperature correction per 334.80 Ampacity, but the amount that you would have to correct by would depend on the ambient temperature, which you have not provided.
If they were in a raceway, which I don't know why you would do that, less a few specific circumstances,
even though the number of CCC's in an individual cable does NOT exceed 3, that would still qualify as more than three CCC's in a raceway and require adjustment per 310.15(B)(3)(a).
No, you would have 8, but the same factor of 0.70 would apply.
Is this a joke because I'm having a DISCUSSION with some mods as opposed just downloading their answers that you feel the need to joke about supplying me with an inaccurate count? Because it's really not funny or clever.
Per 334.80 Ampacity you would have to set their ampacity per the 60*C column at 20A and derate from the 90*C column at 30A "provided the final derated ampacity does not exceed that of a 60*C conductor," which is eerily similar to what we're discussing here and the only reason I'm taking interest in this question.
So 30A * 0.70 * X = Y | X = ambient temperature correction factor; Y = final adjusted and corrected ampacity
Going with the highest 2% design temperature from Copper.org for Tennessee, which is 97*F, and assuming we take a value from the 60*C column of T310.15(B)(2)(a), which I'm honestly not sure if we should do, because I believe the actual conductors within NM are THHN 90*C conductors, but 334.80 requires us to treat NM like a 60*C conductor, then we have a ambient temperature correction of 0.82
So 30A * 0.70 * 0.82 = 17.22A, which satisfies the requirements of 334.80 Ampacity.
If you DID meet the requirements of 240.4(B), then yes, 20A breaker.
If you DID NOT meet the requirements of 240.4(B), then no, 15A breaker.
How'd I do?
Please don't bother addressing each portion of this response, I would like to finish have a discussion with the rest of the people in the thread
@wwhitney and the others. We're making progress and this is just throwing a wrench in the gears.