9. Wang Qiang has four colors of small cylindrical rods, Table 1 lists the lengths of the rods of different colors.
Table 1
\begin{tabular}{|c|c|c|c|c|}
\hline Color & Green & Yellow-Red & Purple & Red \\
\hline Length & $3 \mathrm{~cm}$ & $4 \mathrm{~cm}$ & $8 \mathrm{~cm}$ & $9 \mathrm{~cm}$ \\
\hline
\end{tabular}

Now, several small rods are to be taken and joined together to form a long rod of length $2010 \mathrm{~cm}$, and each of the four colors of small rods must be used at least 81 times. Then the number of different ways to do this is.
9. 91.

From the problem, we have
$$
\begin{array}{l}
3(81+a)+4(81+b)+8(81+c)+9(81+d) \\
=2010,
\end{array}
$$

where $a, b, c, d$ are non-negative integers.
$$
\begin{array}{l}
\text { Simplifying, we get } 3(a+3 d)+4(b+2 c)=66 . \\
\text { Therefore, }(a+3 d, b+2 c) \\
=(4 i+2,15-3 i)(i=0,1, \cdots, 5) . \\
\text { When } a+3 d=4 i+2 \text {, }(a, d) \text { has }\left[\frac{4 i+2}{3}\right]+1
\end{array}
$$

possible values;
$$
\text { When } b+2 c=15-3 i \text {, }(b, c) \text { has }\left[\frac{15-3 i}{2}\right]+1
$$

possible values.
Therefore, the total number of different combinations is
$$
1 \times 8+3 \times 7+4 \times 5+5 \times 4+7 \times 2+8 \times 1=91 \text {. }
$$