jmc

Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following: $$\begin{array}{rl}3a+4b+6c& =41\\ a+b+c& =10\end{array}$$ Subtracting 3 times the second from the first gives $b+3c=11$, or $(b,c)=(2,3),(5,2),(8,1),(11,0)$. The last doesn't work, obviously. This gives the three solutions $(a,b,c)=(5,2,3),(3,5,2),(1,8,1)$. In terms of choosing which goes where, the first two solutions are analogous. For $(5,2,3),(3,5,2)$, we see that there are $2\cdot \frac{10!}{5!2!3!}=10\cdot 9\cdot 8\cdot 7$ ways to stack the crates. For $(1,8,1)$, there are $2\left(\genfrac{}{}{0ex}{}{10}{2}\right)=90$. Also, there are $3^{10}$ total ways to stack the crates to any height. Thus, our probability is $\frac{10\cdot 9\cdot 8\cdot 7+90}{3^{10}}=\frac{10\cdot 8\cdot 7+10}{3^8}=\frac{570}{3^8}=\frac{190}{3^7}$. Our answer is the numerator, $\boxed{190}$.