Total number of $4$ digit numbers whose product of digits is $72$ 
Total number of $4$ digit numbers whose product of digits is $72$

$\bf{My\; Try::}$ Here the possible factor of $72 = 2^{3}\cdot 3^2$
Now here we divide $2^3\cdot 3^3$ into product of $4$ numbers
Like $1\cdot 1\cdot 8\cdot 9$
and $1\cdot 2\cdot 4\cdot 9$
and $1\cdot 2\cdot 6\cdot 6$
and $1\cdot 3\cdot 4\cdot 6$
and $1\cdot 3\cdot 3\cdot 8$
and $2\cdot 2\cdot 3\cdot 6$
and $2\cdot 2 \cdot 2\cdot 9$
So arrangement of these numbers is $$\frac{4!}{2!}+4!+\frac{4!}{2!}+4!+\frac{4!}{2!}+\frac{4!}{2!}+\frac{4!}{3!} = 12+12+12+24+24+12+4=100$$
But answer given is $=112$
So Where i have make mistake, Is there is any other better method, 
Then plz explain here
Thanks 

$1\cdot 1\cdot 8\cdot 9$
and $1\cdot 2\cdot 4\cdot 9$
and $1\cdot 2\cdot 6\cdot 6$
and $1\cdot 3\cdot 4\cdot 6$
and $1\cdot 3\cdot 3\cdot 8$
and $2\cdot 2\cdot 3\cdot 6$
and $2\cdot 2 \cdot 2\cdot 9$
and $2\cdot 3 \cdot 3\cdot 4$
So arrangement of these numbers is $$\frac{4!}{2!}+4!+\frac{4!}{2!}+4!+\frac{4!}{2!}+\frac{4!}{2!}+\frac{4!}{3!}+\frac{4!}{2!} = 12+12+12+24+24+12+4+12=112$$
