<s>[INST] How many ways can we choose $20$ letters from a collection of $9$ b’s , $8$ h’s, $10$ s’s? How many ways can we choose $20$ letters from a collection of $9$ b’s , $8$ h’s, $10$ s’s?
My solution: 
$$\frac{27P20}{8!\times9!\times10!}$$
Is my solution true??
 [/INST] If we put
$$
\begin{gathered}
  P(x,h,b,s) = \left( {1 + hx} \right)^8 \left( {1 + bx} \right)^9 \left( {1 + sx} \right)^{10}  =  \hfill \\
   =  \cdots  + c_{20} x^{20}  +  \cdots  \hfill \\ 
\end{gathered} 
$$
we will have that the coefficient of $x^{20}$ will be composed of all the possible 
combinations of $h,b,s$ such that their exponents sum to $20$ and $h\leq 8,\; b\leq 9\;s\leq10$.
$$
c_{20} (h,b,s) = \left[ {x^{20} } \right]P(x) = \underbrace {120h^8 b^9 s^3  +  \cdots  + 8hb^9 s^{10} }_{\text{36}\;\text{terms}}
$$
So, the number of terms in $c_{20} (h,b,s)$ will count the combinations
which differ by the number of $h,b,s$ (i.e., not considering order).
Unfortunately, there is not a general formula to calculate them when the limits/exponents ($8,9,10$ in this case) are not the same,
if not for the elegant "trick" exposed by @drhab.  
Instead, if order is also considered, then the total number of combinations will be given by
$$
\begin{gathered}
  c_{20} (1,1,1) = \left[ {x^{20} } \right]\left( {1 + x} \right)^8 \left( {1 + x} \right)^9 \left( {1 + x} \right)^{10}  =  \hfill \\
   = \left[ {x^{20} } \right]\left( {1 + x} \right)^{27}  = \left( \begin{gathered}
  27 \\ 
  20 \\ 
\end{gathered}  \right) = \left( \begin{gathered}
  27 \\ 
  7 \\ 
\end{gathered}  \right) \hfill \\ 
\end{gathered} 
$$
</s>