Combinatorics : Choosing r elements from a multiset of n elements, given some elements in multiset are identical to each other Given a multiset = {a, a, a, b, b, b, b, c, c}. The multiset has:
number of a's = 3
number of b's = 4
number of c's = 2
In how many ways we can select any 3 elements from this multiset?

My solution to this question:
to get the required number of ways in which we can select three elements from that multiset we need to find the coefficient of $x^3$ in $\{\left( 1+x+x^2+x^3 \right) \left( 1+x+x^2+x^3 +x^4 \right) \left( 1+x+x^2 \right)\}$
Now, I'm stuck. How to proceed with this? Do I have to manually sum the number of $x^3$ we can obtain after multiplication, to get the coefficient.
Can this be done using Generating functions? If yes, then how?
Give references to book/study material online link, if you want to escape writing down complete solution.

Here the numbers are so small that it’s not hard simply to write out the terms of degree $3$ and count them:
$$\begin{align*}
&x^3\cdot1\cdot1\\
&x^2\cdot1\cdot x\\
&x^2\cdot x\cdot 1\\
&x\cdot 1\cdot x^2\\
&x\cdot x\cdot x\\
&x\cdot x^2\cdot 1\\
&1\cdot x\cdot x^2\\
&1\cdot x^2\cdot x\\
&1\cdot x^3\cdot 1
\end{align*}$$
However, counting multisets of a given cardinality isn’t too hard a problem theoretically. If we had at least $3$ of each letter available, the answer would be simply
$$\left(\!\!\binom{3}{\color{red}3}\!\!\right)\;,$$
the number of multisets of cardinality $3$ taken from a set of $3$ available types of object. By a standard stars and bars argument this is equal to
$$\binom{3+\color{red}3-1}{\color{red}3}=\binom53=10\;.$$
However, we have only $2$ $c$s available, so we have to throw away any multiset with more than $2$ $c$s. Since we’re looking at multisets of size $3$, there is only one, and we get $10-1=9$ as the final count.
You can also work directly with your generating function. Note that
$$1+x+x^2+x^3=\frac{1-x^4}{1-x}\;,$$
and similarly for the other two factors, so your generating function can be rewritten as
$$\frac{(1-x^3)(1-x^4)(1-x^5)}{(1-x)^3}\;.\tag{1}$$
Now
$$\begin{align*}
(1-x^3)(1-x^4)(1-x^5)&=(1-x^3-x^4+x^7)(1-x^5)\\
&=1-x^3-x^4-x^5+x^7+x^8+x^9-x^{12}\;,
\end{align*}$$
so $(1)$ is a sum of terms of the form $\dfrac{x^m}{(1-x)^3}$. A useful generating function to know is
$$\frac1{(1-x)^3}=\sum_{n\ge 0}\binom{n+2}nx^n=\sum_{n\ge 0}\binom{n+2}2x^n\;,$$
from which we get
$$\frac{x^m}{(1-x)^3}=\sum_{n\ge 0}\binom{n+2}2x^{m+n}\;.\tag{2}$$
The coefficient of $x^3$ in $(2)$ is 
$$\binom{(3-m)+2}2=\binom{5-m}2\;,$$
which is $0$ for $m>3$, so we need only deal with $\dfrac1{(1-x)^3}-\dfrac{x^3}{(1-x)^3}$, with $m=0$ and $m=3$, getting
$$\binom52-\binom22=9$$
as before.
