\[
a+b+c+d=27,\qquad 1\le a\le9,\;0\le b,c,d\le9
\]
\[
\text{Let }x_1=a-1,\;x_2=b,\;x_3=c,\;x_4=d.
\]
\[
x_1+x_2+x_3+x_4=26,\qquad 0\le x_1\le8,\;0\le x_2,x_3,x_4\le9
\]
\[
\text{Total nonnegative solutions: }\binom{26+4-1}{3}=\binom{29}{3}=3654
\]
\[
\begin{aligned}
\text{Subtract violations:}&\\
x_1\ge9:&\quad \binom{17+3}{3}=\binom{20}{3}=1140,\\
x_2\ge10\ (\text{or }x_3\ge10,\ x_4\ge10):&\quad \binom{16+3}{3}=\binom{19}{3}=969\ (\text{each})
\end{aligned}
\]
\[
\text{Sum of single violations }=1140+3\cdot969=4047
\]
\[
\text{Double violations:}
\]
\[
\begin{aligned}
x_1\ge9,\;x_i\ge10\ (i=2,3,4):&\quad \binom{7+3}{3}=\binom{10}{3}=120\ (\text{each, }3\ \text{cases})\\
x_i\ge10,\;x_j\ge10\ (2\le i<j\le4):&\quad \binom{6+3}{3}=\binom{9}{3}=84\ (\text{each, }3\ \text{cases})
\end{aligned}
\]
\[
\text{Sum of double violations }=3\cdot120+3\cdot84=360+252=612
\]
\[
\text{No triple violations (sum would be negative).}
\]
\[
\text{Valid numbers}=3654-4047+612=219
\]
\[
\boxed{219}
\]
\end{document}
