Solving a combinatorics problem through a linear equation. Is my solution correct? Are there any other ways to solve it? Problem:
Find the number of ways that we can assign 20 projects to 15 people, such that each person will get assigned at least one project
My solution:
The above problem can be coded into finding the integer solutions of this linear equation:
$x_1 + x_2 + x_3 +... + x_{15} = 20$
Since each person should get assigned at least one project, every $x_i>=1$ so $x_i - 1 >= 0$. If we set $y_i = x_i - 1$, we have the equation
$y_1 + y_2 +...+y_15 = 20 = (x_1 - 1) + (x_2 -1) +...+(x_{15} -1) = x_1 +x_2 +...+x_{15} = 20-15=5$
So the above equation can be solved as $\binom {15+5-1}5$
Is this correct? And are there any other different approaches to this problem?
 A: If the projects are $\color{blue}{\text{identical}}$ , then your approach is $\color{blue}{\text{correct}}$!. For a second way  , generating functions can be good way such that find the coefficient of $x^{20}$ in the expansion of $(\frac{1-x^{16}}{1-x} -1 )^{15}$ , we limited the generating function until $x^{15}$ , because one can take at most $15$ projects if each one take at least once.
$\mathbf{\text{NOTE:}}$ If we do not restrict the exponential generating function , it will be also okey.However , the calculation by hand can be easier when you restrict it.
Calculation of generating function
As you see $[x^{20}]=11628 = C(19,5)$
$\mathbf{\text{NOTE 2:}}$ It is not stated that whether the projects identical or not , so if they are $\color{blue}{\text{distinct}}$ , then use exponential generating function. If everyone gets at least once , then $$x + \frac{x^2}{2!}  + \frac{x^3}{3!} +\frac{x^4}{4!}+... = (e^x -1)$$ (Remember the Taylor expansion from Calculus years)
Hence , find the coefficient of $\frac{x^{20}}{20!}$ in $(e^x - 1)^{15}$ or find the coefficient of $x^{20}$ and multiply it by $20!$
Calculation of exponentials
So , $$20! \times \frac{1945}{8}=591,499,300,737,945,600,000$$
