The product of four digits is $336$.

Obviously one must be seven so the product of three digits is $48$.

Listing factor pairs of $48$:

$(1,48)$ --> from here we can make the $48$ into $6,8$, so the pairs $(1,6,8)$ appear.

$(2,24)$ --> from here we can make $24$ into $8,3$ or $4,6$ to make $(2,4,6)$ and $(2,3,8)$

$(4,12)$ --> from here we can make $12$ into $2,6$ or $3,4$ so $(4,2,6)$, $(4,3,4)$

$(6,8)$ --> we can either change the $6$ into $1,6$ or $2,3$ or change $8$ into $1,8$ or $2,4$ to make $(6,1,6), (6,2,4), (1,6,8), (2,3,8)$.

So we can have $(1,6,8), (2,4,6), (2,3,8), (4,2,6), (4,3,4), (6,1,6), (6,2,4), (1,6,8), (2,3,8)$. We double counted a few, so lets take some out:

$(1,6,8), (2,4,6), (2,3,8), (4,3,4), (6,1,6)$.

So we can have $(7,1,6,8), (7,2,4,6), (7,2,3,8), (7,4,3,4), (7,6,1,6)$. Now, the first three pairs can be arranged in $4!=24$ ways. The last two can be arranged in $\frac{4!}{2!}=12$ ways. The answer is $24*3+12*2=24*4=\boxed{96}$.

Whew!

This is what I would do in contest to eliminate silly errors, but you can obviously just list triples of $48$ without my process.