jmc

First, we find how many ways are there to write 9 as the sum of 1s, 2s and 4s, where the order of the addends does not matter. We find these cases: $$\begin{array}{rl}& 4+4+1\\ & 4+2+2+1\\ & 4+2+1+1+1\\ & 4+1+1+1+1+1\\ & 2+2+2+2+1\\ & 2+2+2+1+1+1\\ & 2+2+1+1+1+1+1\\ & 2+1+1+1+1+1+1+1\\ & 1+1+1+1+1+1+1+1+1\end{array}$$There are $3!/2!=3$ distinguishable orders for the first sum, $4!/2!=12$ for the second sum, $5!/3!=20$ for the third sum, $6!/5!=6$ for the fourth sum, $5!/4!=5$ for the fifth sum, $6!/3!3!=20$ for the sixth sum, $7!/5!2!=21$ for the seventh sum, $8!/7!=8$ for the eighth sum, and $1$ for the last sum. In total, there are $\boxed{96}$ distinguishable ways are there to write $9$ as the sum of $1\text{’s},$ $2\text{’s}$ and $4\text{’s}.$