Distribution of 10 different books to three students. Could someone confirm my solution to this combinatoric question?
Question:

In how many ways can 10 different books be distributed to three
  students so that each student receives at least three books?

Solution:
There are three cases: first student gets four books, second student gets four books, or third student gets four books.
Case 1:


*

*Pick four books out of the available 10 books and give them to the first student, $\dbinom{10}{4}$.

*Then, pick three books out of the remaining six books and give them to the second student, $\dbinom{6}{3}$.

*Lastly, give the remaining three books to the last student, $\dbinom{3}{3}$.


Case 2:


*

*Pick three books out of the available 10 books and give them to the first student, $\dbinom{10}{3}$.

*Then, pick four books out of the remaining seven books and give them to the second student, $\dbinom{7}{4}$.

*Lastly, give the remaining three books to the last student, $\dbinom{3}{3}$.


Case 3:


*

*Pick three books out of the available 10 books and give them to the first student, $\dbinom{10}{3}$.

*Then, pick three books out of the remaining seven books and give them to the second student, $\dbinom{7}{3}$.

*Lastly, give the remaining four books to the last student, $\dbinom{4}{4}$.


Answer: $P = \dbinom{10}{4}\cdot\dbinom{6}{3}\cdot\dbinom{3}{3} + \dbinom{10}{3}\cdot\dbinom{7}{4}\cdot\dbinom{3}{3} + \dbinom{10}{3}\cdot\dbinom{7}{3}\cdot\dbinom{4}{4}$
 A: To summarize what was said above in the comments:
We begin by noticing that exactly one of the three students must receive four books while the other two students will each receive three books.  No other arrangements are possible.
(we must account for all possibilities in terms of who receives the fourth book, it will not always be a specific person)
Approach via multiplication principle:


*

*Pick which of the students receives the four books: $3$ possibilities

*Pick which four books said student receives:  $\binom{10}{4}$ possibilities

*Pick which three books the shortest remaining student receives: $\binom{6}{3}$  (or some other convenient way to have ordered the students, such as name or age or studentid#)

*Pick which three books the final student receives: $\binom{3}{3}$ possibilities


There are thus $3\binom{10}{4}\binom{6}{3}\binom{3}{3} = \frac{3\cdot 10!}{4!3!3!}$ possible outcomes.
The answer in the current version of the post is now correct and is equal to the answer I give in the previous line.

In attempting to answer the same problem but with a larger number of books to distribute, consider approaching via Inclusion-Exclusion.
If there are $n>9$ books to distribute, and each student must receive at least $3$ books, let the students be labeled $a,b,c$.  Let $A$ represent the event that student $a$ does not receive at least three books.  Similarly, events $B$ and $C$ will be for students $b$ and $c$ not receiving at least three books respectively.
We are tasked with finding
$$(\star)~~|A^c\cap B^c\cap C^c| = |\Omega\setminus (A\cup B\cup C)| = |\Omega|-|A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|$$
where $\Omega$ represents the number of ways of distributing the books regardless of restrictions.
$|\Omega| = 3^n$ (since it can be thought of as the number of functions from $\{1,2,\dots n\}\to \{a,b,c\}$)
$|A|$ is a bit more challenging to count.  Break into cases: $a$ receives no books, $a$ receives $1$ book, $a$ receives $2$ books.  In each case, decide which books $a$ receives, and distribute the rest among $b$ and $c$.
$|A|=2^n + n2^{n-1} + \binom{n}{2}2^{n-2}$.  By symmetry, this is also the value of $|B|$ and $|C|$
$|A\cap B|$ will also be tedious to calculate: the following nine cases are possible: $a$ and $b$ receive zero, $a$ receives $1$ while $b$ receives zero, ... $a$ receives $2$ while $b$ receives $2$.  In each case, $c$ will receive the rest of the books.
$|A\cap B| = 1+n+\binom{n}{2}+n+n(n-1)+\binom{n}{2}(n-3)+\binom{n}{2}+n\binom{n-1}{2}+\binom{n}{2}\binom{n-2}{2}$.  By symmetry, this is also the value of $|A\cap C|$ and $|B\cap C|$
$|A\cap B\cap C|$ is equal to zero in this case, since it is impossible for all three to receive too few books.
Combining all of this information into equation $(\star)$ will give an answer.
If you wish to, I'm sure by plugging in $n=10$, the answer will match the one given earlier in terms of value (though it may look much more tedious).  Despite it looking more tedious for the case of $n=10$, it will be much cleaner than what you would see for a case such as $n=100$ had you approached directly via cases.
