List of Three Letters I'm having issues with my logic on this problem:

How many ways are there to form a list of three letters from the letters in the word COMBINATORICS if the letters cannot be used more often than they appear in COMBINATORICS? 

I'm trying to think of this as a set with 13 elements (2 Cs, 2 Os, 2 Ms, etc).
Thus, 13*12*11 would give us 1716 possibilities but apparently that isn't correct.
Any help or clarifications would be greatly appreciated.

A general method for solving this kind of problem uses exponential generating functions.  Let
$$
\begin{aligned}
G(x,C,O,I,M,\ldots)=&\left(1+\frac{Cx}{1!}+\frac{C^2x^2}{2!}\right)\left(1+\frac{Ox}{1!}+\frac{O^2x^2}{2!}\right)\left(1+\frac{Ix}{1!}+\frac{I^2x^2}{2!}\right)\left(1+\frac{Mx}{1!}\right)\\
&\times\left(1+\frac{Bx}{1!}\right)\left(1+\frac{Nx}{1!}\right)\left(1+\frac{Ax}{1!}\right)\left(1+\frac{Tx}{1!}\right)\left(1+\frac{Rx}{1!}\right)\left(1+\frac{Sx}{1!}\right).
\end{aligned}
$$
The variables $C,$ $O,$ $I,$ $M,$ and so on, are there just to indicate which factor counts which letter, and may all be set equal to $1.$  If we do this, we get
$$
G(x):=G(x,1,1,1,1,\ldots)=\left(1+\frac{x}{1!}+\frac{x^2}{2!}\right)^3\left(1+\frac{x}{1!}\right)^7
$$
The coefficient of $x^3$, multiplied by $3!,$ will be the desired number of arrangements.
We get
$$
\begin{aligned}
G(x)&=\left((1+x)^3+\binom{3}{1}\frac{x^2}{2!}(1+x)^2+\binom{3}{2}\left(\frac{x^2}{2!}\right)^2
(1+x)+\binom{3}{3}\left(\frac{x^2}{2!}\right)^3\right)\,\left(1+x\right)^7\\
&=(1+x)^{10}+\binom{3}{1}\frac{x^2}{2!}(1+x)^9+\ldots.
\end{aligned}
$$
The coefficient of $x^3$ in this expression is $\binom{10}{3}+\binom{3}{1}\frac{1}{2!}\binom{9}{1}.$  Multiplying by $3!$ gives $_{10}P_3+3\cdot\binom{3}{1}\binom{9}{1}.$
