<s>[INST] Find the number of solutions of $a_0+a_1+a_2=17$ if $2\le a_0\le 5$, $3\le a_1\le 6$, $4\le a_2\le 7$. 
Find the number of solutions of $a_0+a_1+a_2=17$ if $2\le a_0\le 5$,
  $3\le a_1\le 6$, $4\le a_2\le 7$.

This is an exmaple given by my professor, and his solution is:  

The number of solutions to this equation satisfying the given
  constraints is equal to the coefficient of $x^{21}$ in the expression
  $$\left(\sum_{n=2}^5x^n\right)\left(\sum_{n=3}^6x^n\right)\left(\sum_{n=4}^7x^n\right)=x^9\left(\sum_{n=0}^3x^n\right)\left(\sum_{n=0}^3x^n\right)\left(\sum_{n=0}^3x^n\right).$$
  This must be 3, since the coefficient of $x^8$ in $(1+x+x^2+x^3)^3$ is
  3.

I understand actually nothing here. The expression of summation just pops up, and then comes the answer. What I only know that the idea of generating function is used here.
Can someone help explain how the steps in the solution come?
 [/INST] generating function.
This is another method that you may be interested.
$2\le a_0\le 5$, $3\le a_1\le 6$, $4\le a_2\le 7$.  
$0\le a_0-2\le 5-2$, $0\le a_1-3\le 6-3$, $0\le a_2-4\le 7-4$.  
$0\le b_0\le 3$, $0\le b_1\le 3$, $0\le b_2\le 3$.  
$ a_0-2 = b_0, a_1-3 = b_1, a_2-4 = b_1$
$ a_0 =b_0+2,  a_1 = b_1+3, a_2 = b_2+4$
$b_0+b_1+b_2+2+3+4 = 17$
$b_0+b_1+b_2 = 17-9 = 8$
Now $y_0 = 3-b_0, y_1 = 3-b_1, y_2= b_2 = 3-b_2, y_i's \gt 0$
$9-y_0+y_1+y_2 = 8$
$ y_0+y_1+y_2 = 1$
The number of solutions  $= {(1+3-1)\choose(3-1)} = {3\choose2} = 3$
What I have given you is another solution using stars and bars.
$(x^2+x^3+x^4+x^5)(x^3+x^4+x^5+x^6)(x^4+x^5+x^6+x^7)$ is what your professor has shortened with summation signs.  
How did you get this, the first expresssion in the bracket corresponds to values that $a_0$ can take. The way it is represented in generating function is $a_0=0$ is 1, $a_0=1$ is $x$, $a_0=2$ is $x^2$ and so on.  $a_0$ can take values of {2,3,4,5} hence the first four terms of $x$ in the first bracket.  Next a_1=3 is $x^3$, $a_1=4$ is $x^4$. $a_1=5$, is $x^5$ and $a_1=6$ is $x^6$.  and hence the first four terms of x in the second bracket and now you can figure out how the thrid bracket of x terms are derived.
Now take $x^2$, $x^3$ and $x^4$ from all three bracketed expresssions what you then have is
$$x^2(1+x+x^2+x^3)\cdot x^3(1+x+x^2+x^3)\cdot x^4(1+x+x^2+x^3)=x^9(1+x+x^2+x^3)^3$$
What you are looking for is the coefficient of $x^{17}$, since we already took out $x^9$, we are looking for the coefficient of $x^8$ which is second term of the expansion of the expression in the brackets.  if you expand $(1+x+x^2+x^3)^3$ using the binomial theorem, the equation becomes
$$(1+x+x^2+x^3)^3=\sum_{n=0}^3\binom{n}{3}1^n(x+x^2+x^3)^{3-n}$$
There can't be $x^8$ in $(x+x^2+x^3)^0$, $(x+x^2+x^3)$, and $(x+x^2+x^3)^2$, so $x^8$ appears only in $(x+x^2+x^3)^3$. Note that
$$(x+x^2+x^3)^3=(x+x^2+x^3)^2\cdot (x+x^2+x^3)=(x^6+2x^5+\dots)(x+x^2+x^3)$$
Hence, we obtain that the coefficient of $x^8$ is 3.
</s>