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 $_{10}P_3+3\cdot\binom{3}{1}\binom{9}{1}.$