Selecting 180 days from 366: the probability of even distribution across months, or not having September among the first 30 
In a draft lottery containing the 366 days of the year (including February 29). Select 180 days (draw 180 without replacement).
a) What is the probability that the 180 days drawn are evenly distributed among the twelve months?
b) What is the probability that the first 30 days drawn contain none from September?

I understood part a). But they say the answer for part b is:
Number of combinations of 336 taken 30 at a time / Number of combinations of 366 taken 30 at a time.
I understand the bottom but not the top.
When you take the number of combinations of 336 taken 30 at a time how can you make sure
you are letting out the 30 days from September and not other 30 days from any other month.
What would the top be if the question said: What is the probability that the first 30 days drawn contain none from December? The same top?
I do not think so. It is less likely for December than for September because Dec has 31 days.
I would appreciate your comments regarding my concern. Thank you very much in advance.
 A: (a)
$$ \displaystyle \frac{{31 \choose 15}{29 \choose 15}...{31 \choose 15}}{366 \choose 180}  $$
(b)
$$ {30 \choose 0}{336 \choose 180}\over  {366 \choose 180}   $$
Hypergeometric distribution. We divide the year up into different categories: 12 months in  (a)  and September versus the rest of the year in (b).
In both problems note how the sum across the "rows" is the same in the numerator and denominator:
31+29+...+31=366
15+15+...15=180.
30+336=366
0+180=180
This helps you make sure you've accounted for everything.
A: The September part, besides informing us that we are talking about a group of 30 days, is irrelevant. We are not "sure" that we are taking out the 30 days from September, but the calulcation remains the same no matter which group of 30 days we are talking about.
Remember, we are not sure that anything happens, we are only calculating the probability that certain things happen. Just as the probability of throwing a die and getting a one is $\frac{1}{6}$, the probability of getting a six is also $\frac{1}{6}$. It doesn't matter whether we're talking about ones or sixes, the important thing is the possible number of outcomes, and the outcomes that we are interested in.
For December, it would have to be altered to account for the fact that December has 31 days, not 30, as you say, and then similar calculations take place.
