Probability: Problem with $30$ books distributed to $3$ libraries 
We have $30$ books. $15$ of them are Literature Books(LB) and the other $15$ are Mathematic Books(MB). We randomly distribute those $30$ books to $3$ libraries, giving to each library $10$ books. What's the probability that the first library gets $10$ LB and the second one gets $5$ LB?

This is a problem from Probability class I am attending. I am a bit confused on this...i know there are $\binom{30}{10}$ ways to distribute $10$ books from $30$ to one library but i have to take into account what's happening to the other $2$ libraries, plus there are $\binom{15}{10}$ ways to give to the first library $10$ LB. And now comes my confusion on finding the rest of these and combining them. Even a small hint may help me solve it so thanks in advance.
 A: After you give the first library 10 books, you have $\binom{20}{10}$ ways to choose $10$ books of the remaining $20$ to give to the second library.  There no choice for the third library -- it gets whatever's left.  So, there are $$\binom{30}{10}\binom{20}{10}$$ ways to distribute the books.
Now apply the same kind of reasoning to the given distribution.  After giving $10$ LB to the first library, how many ways are there to give the second library $5$ LB?
For this particular problem, there's a shortcut.  What does the third library get?  
A: First the probability that $10$ literature books are distributed to the first libary and no mathematic books to the second libary:
$$\frac{\binom{15}{10}\cdot \binom{15}{0}}{\binom{30}{10}}$$
Now we can distribute 5 literature books to the second libary. That means that 5 of 15 mathematic books are distributed to the second libary.
$$\frac{\binom{5}{5}\cdot \binom{15}{5}}{\binom{20}{10}}$$
The remaining books can be distributed arbitrarily, since all conditions are fullfilled
