I'm trying to solve the following problem:

In how many ways can you distribute 100 identical balls among 3 different boxes (box A, box B and box C), such that each box will get at least 20 and at most 50 balls?

I'm trying to solve it using a generating function. 
Please help me check the correctness. And please let me know if there's an easier way to solve this problem.

So I'm trying to find the integer solution to:

$c\_1+c\_2+c\_3=100$ with c1, c2, c3 varying from 20 to 50 balls.

The associated polynomial associated with the given equation is:
$(x^{20}+x^{21}+x^{22}+...+x^{50})^3$

I think the solution to the problem is the coefficient of $x^{100}$ in the equation $(x^{20}+x^{21}+x^{22}+...+x^{50})^3$, so I'm trying to find the coefficient of $x^{100}$.

$(x^{20}+x^{21}+x^{22}+...+x^{50})^3={(x^{20})}^3 {(1+x+x^2+...+x^{30})}^{3}$
$= {(x^{60})} {(1+x+x^2+...+x^{30})}^{3}$

Using the geometric series, this becomes:

${(x^{60})} { ( \frac{1-x^{31} }{1-x} ) }^{3}$
$= {(x^{60})} { ( 1 -3x^{31}+3x^{62}-x^{93} ) }{(1-x)}^{-3}$

Next I'm trying to find ways the coefficients of $x^{100}$ can be obtained:

$x^{60}x^{0}x^{40}$

$x^{60}x^{31}x^{9}$

Then I'm trying to find the value of the coefficient of $x^{100}$ by summing the values of these coefficients:

$1\*1 \* {3+40-1 \choose 40} + 1\*-3\*{3+9-1 \choose 9}$
$= {42 \choose 40} -3 \*{11 \choose 9}$ 
$= 861 - 3 \* 55 = 696$