<s>[INST] The number of ways to write $10$ as the sum of five natural numbers not equal to $3$ How many answers are there for the equation
$$x_1+x_2+x_3+x_4+x_5=10$$
given that $x_1,x_2\dots x_5\in\Bbb{Z^{0+}}\setminus\{3\}$.
 [/INST] The generating function for the number of ways to sum to $k$ with these numbers is
$$
\begin{align}
&\left(\frac1{1-x}-x^3\right)^5\\
&=\frac1{(1-x)^5}-\frac{5x^3}{(1-x)^4}+\frac{10x^6}{(1-x)^3}-\frac{10x^9}{(1-x)^2}+\frac{5x^{12}}{1-x}-x^{15}\\
&=\sum_{k=0}^\infty\left[\binom{k+4}{k}x^k-5\binom{k+3}{k}x^{k+3}+10\binom{k+2}{k}x^{k+6}\right]\\
&+\sum_{k=0}^\infty\left[-10\binom{k+1}{k}x^{k+9}+5\binom{k}{k}x^{k+12}-\binom{k-1}{k}x^{k+15}\right]\\[3pt]
&\small=\sum_{k=0}^\infty\left[\binom{k+4}{k}-5\binom{k}{k-3}+10\binom{k-4}{k-6}-10\binom{k-8}{k-9}+5\binom{k-12}{k-12}-\binom{k-16}{k-15}\right]x^k
\end{align}
$$
Thus, the number of ways to solve $x_1+x_2+x_3+x_4+x_5=k$ where each $x_j$ is a non-negative integer not equal to $3$ is
$$
\small\binom{k+4}{k}-5\binom{k}{k-3}+10\binom{k-4}{k-6}-10\binom{k-8}{k-9}+5\binom{k-12}{k-12}-\binom{k-16}{k-15}
$$
Setting $k=10$ gives
$$
\binom{14}{10}-5\binom{10}{7}+10\binom{6}{4}-10\binom{2}{1}=531
$$
Note that the second to last term is $0$ for $k\lt12$ and the last term is $0$ for $k\ne15$.
</s>