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??
 A: You are asked to find the number of sums $b+h+s=20$ where $b,h,s$ are nonnegative integers with $b\leq9$, $h\leq8$ and $s\leq10$.
Define $b'=9-b$, $h'=8-h$ and $s'=10-s$.
Then the question comes to the same as finding the number of sums $b'+h'+s'=7$ where again $b',h',s'$ are nonnegative integers with $b'\leq9$, $h'\leq8$ and $s'\leq10$.
Fortunately the condition that $b',h',s'$ are nonnegative integers with $b'+h'+s'=7$ allready implies that $b'\leq9$, $h'\leq8$ and $s'\leq10$, so that the last condition can be put aside now.
Then with stars and bars we find $\binom{7+2}{2}=36$ possibilities.
The trick I used here does not work always, but it is a good habit to check that out.
A: The solution above is very nice. I add the brute force method from my comment above if it is of interest to anyone. You can have 9 b's, 8 h's, and 3 s's which we will denote by 9 8 3. The list of possible combinations is:
9 8 3, 9 7 4, ..., 9 1 10 (8 of these)
8 8 4, 8 7 5, ..., 8 2 10 (7 of these)
7 8 5, 7 7 6, ..., 7 3 10 (6 of these)
...
...
3 8 9, 3 7 10 (only 2)
2 8 10 (only 1)
So the total is 8+7+6+5+4+3+2+1=$\frac {8\cdot 9} 2=36$
A: 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
$$
\begin{gathered}
  c_{20} (1,1,1) = \left[ {x^{20} } \right]\left( {1 + x} \right)^8 \left( {1 + x} \right)^9 \left( {1 + x} \right)^{10}  =  \hfill \\
   = \left[ {x^{20} } \right]\left( {1 + x} \right)^{27}  = \left( \begin{gathered}
  27 \\ 
  20 \\ 
\end{gathered}  \right) = \left( \begin{gathered}
  27 \\ 
  7 \\ 
\end{gathered}  \right) \hfill \\ 
\end{gathered} 
$$
