Number of non-negative integer solutions of $41x_0 + x_1+x_2+...+ x_k=41n$ Number of non-negative integer solutions of $41x_0 + x_1+x_2+...+ x_k=41n$
I've attempted this question by doing cases, such as when $x_0 = 0$, we get $\binom{41n + k - 1}{k-1}$ solutions. When $n=1$, we have $\binom{41(n-1) + k - 1}{k-1}$ 
Is this the right approach?

Using the "stars and bars" bijection, you know that the number of non-negative integer solutions to
$$
x_1+\cdots+x_k=m
$$
is $\binom{m+k-1}{k-1}$. Now, apply this in turn to $m=41n, 41(n-1), \ldots, 0$ corresponding to all possible choices of $x_0$ (and moving it over to the right side of the equation). Thus the total number of solutions is
$$
\binom{41n+k-1}{k-1}+\binom{41(n-1)+k-1}{k-1}+\cdots+\binom{k-1}{k-1}.
$$
