how to enumerate and index partial permutations with repeats I know how to count permutations of, say, p red balls and q white balls and r blue balls.  I also know how to count permutations of part of a set of distinct objects.  But I don't know an efficient way to count partial permutations with repetition.
To illustrate what I mean: suppose I have a supply of eight As, seven Bs and six Cs, and want to know how many sequences of length 12 can be made from these.  I can do it by adding up the permutations of 8 As and 4 Bs, 8 As and 3 Bs and one C, ... but that's tedious (this example has 45 cases).  Is there a better way?
Finally, is there a procedure to generate the nth such sequence in lexicographic order, without cranking out all of them?

We can count the number of different sequences with the help of exponential generating functions.

Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series we can calculate the wanted number of sequences as
\begin{align*}
12!&[z^{12}](1+\frac{z}{1!}+\frac{z^2}{2!}+\cdots+\frac{z^8}{8!})(1+\frac{z}{1!}+\frac{z^2}{2!}+\cdots+\
\frac{z^7}{7!})\\
&\qquad\cdot(1+\frac{z}{1!}+\frac{z^2}{2!}+\cdots+\frac{z^6}{6!})\\
&=484110
\end{align*}
The calculation was done with the help of Wolfram Alpha.

Note: The result coincides with the calculation of @N.F.Taussig. We obtain
\begin{align*}
\binom{12}{8}2^4&+\binom{12}{7}2^5+\binom{12}{6}2^6+\binom{12}{5}\left(2^7-1\right)\\
&+\binom{12}{4}\left[2^8-2\binom{8}{8}-\binom{8}{7}\right]\\
&+\binom{12}{3}\left[2^9-2\binom{9}{9}-2\binom{9}{8}-\binom{9}{7}\right]\\
&+\binom{12}{2}\left[2^{10}-2\binom{10}{10}-2\binom{10}{9}-2\binom{10}{8}-\binom{10}{7}\right]\\
&+\binom{12}{1}\left[\binom{11}{5}+\binom{11}{6}+\binom{11}{7}\right]\\
&+\binom{12}{0}\left[\binom{12}{6}+\binom{12}{7}\right]\\
&=7920+25344+59136+100584+121770\\
&\qquad+100320+52272+15048+1716\\
&=484110
\end{align*}
