Product of digits of a 7 digit number is $2^43^4$ Find the number of seven digit numbers whose product of digits is $2^43^4$.
One method is to list out all possible sets of seven digits that give this product, and then find number of permutations for each case. But there are too many cases that way.
Is there a shorter method?

I propose to count these numbers with respect to the number of digits 6. 
We have five cases: ZERO, ONE, TWO, THREE, FOUR. In the following formulas we place the $6$s then the powers of $2$ (i. e. $2,4,8$) and finally the powers of $3$ (i. e. $3,9$).
If there are ZERO digits 6 then we have
$$
\binom{7}{4}\left[\binom{3}{4}+3\binom{3}{3}+\binom{3}{2}\right]
+3\binom{7}{3}\left[\binom{4}{4}+3\binom{4}{3}+\binom{4}{2}\right]
\\+
3\binom{7}{2}\left[\binom{5}{4}+3\binom{5}{3}+\binom{5}{2}\right]
=5040.$$
If there is ONE digit 6 then we have
$$\binom{7}{1} \left[ \binom{6}{3}\left[\binom{3}{3}+2\binom{3}{2}\right]
+2\binom{6}{2}\left[\binom{4}{3}+2\binom{4}{2}\right]+
\binom{6}{1}\left[\binom{5}{3}+2\binom{5}{2}\right]
\right]
=5600.$$
If there are TWO digits 6 then we have
$$\binom{7}{2}\left[ \binom{5}{2}\left[\binom{3}{2}+\binom{3}{1}\right]
+\binom{5}{1}\left[\binom{4}{2}+\binom{4}{1}\right]
\right]=2310.$$
If there are THREE digits 6 then we have
$$\binom{7}{3}\binom{4}{1}\binom{3}{1}=420.$$
If there are FOUR digits 6 then we have
$$\binom{7}{4}=35.$$
Finally, the total number is
$$5040+ 5600+2310+420+35=13405.$$
