smc

Note that $96=2^5\cdot 3$. Since there are at most six not necessarily distinct factors $>1$ multiplying to $96$, we have six cases: $k=1,2,...,6.$ Now we look at each of the six cases. $k=1$: We see that there is $1$ way, merely $96$. $k=2$: This way, we have the $3$ in one slot and $2$ in another, and symmetry. The four other $2$'s leave us with $5$ ways and symmetry doubles us so we have $10$. $k=3$: We have $3,2,2$ as our baseline. We need to multiply by $2$ in $3$ places, and see that we can split the remaining three powers of $2$ in a manner that is $3-0-0$, $2-1-0$ or $1-1-1$. A $3-0-0$ split has $6+3=9$ ways of happening ($24-2-2$ and symmetry; $2-3-16$ and symmetry), a $2-1-0$ split has $6\cdot 3=18$ ways of happening (due to all being distinct) and a $1-1-1$ split has $3$ ways of happening ($6-4-4$ and symmetry) so in this case we have $9+18+3=30$ ways. $k=4$: We have $3,2,2,2$ as our baseline, and for the two other $2$'s, we have a $2-0-0-0$ or $1-1-0-0$ split. The former grants us $4+12=16$ ways ($12-2-2-2$ and symmetry and $3-8-2-2$ and symmetry) and the latter grants us also $12+12=24$ ways ($6-4-2-2$ and symmetry and $3-4-4-2$ and symmetry) for a total of $16+24=40$ ways. $k=5$: We have $3,2,2,2,2$ as our baseline and one place to put the last two: on another two or on the three. On the three gives us $5$ ways due to symmetry and on another two gives us $5\cdot 4=20$ ways due to symmetry. Thus, we have $5+20=25$ ways. $k=6$: We have $3,2,2,2,2,2$ and symmetry and no more twos to multiply, so by symmetry, we have $6$ ways. Thus, adding, we have $1+10+30+40+25+6=\boxed{112}$.