Count the numbers in which the product of the digits is 48 (1 not allowed) 
Count the numbers in which the product of the digits is 48 (1 not allowed as digit).

Proposed solution: I addressed the problem by brute force, first noticing that
$$48=2^4\times 3.$$
Then I found the possible groups of digits to be permuted for each number: (1) 2,2,2,2,3; (2) 2,2,2,6; (3) 4,2,2,3; (4) 4,2,6; (5) 4,4,3; (6) 8,2,3; (7) 8,6.
By adding the permutations with repetition for each case, I've found 38 possible numbers. Which agrees with the answer provided in the source.
My problem: I found this solution is not elegant and I'm wondering if there is a better way to address problems like these. By brute force it is easy to undercount terms (as I did as first solving the problem). Any help?
 A: 
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bound: $1\le\text{n}\le10^7$.

I wrote and ran some Mathematica-code:
In[1]:=Clear["Global`*"];
ParallelTable[
  If[TrueQ[Product[
       Part[IntegerDigits[n], k], {k, 1, 1 + Floor[Log10[n]]}] == 48 &&
      MemberQ[IntegerDigits[n], 1] == False], n, Nothing], {n, 1, 
   10^7}] //. {} -> Nothing

Running the code gives:
Out[1]={68, 86, 238, 246, 264, 283, 328, 344, 382, 426, 434, 443, 462, 624, 
642, 823, 832, 2226, 2234, 2243, 2262, 2324, 2342, 2423, 2432, 2622, 
3224, 3242, 3422, 4223, 4232, 4322, 6222, 22223, 22232, 22322, 23222, 
32222}

So, the sum of the numbers is given by $179141$. I found that using:
In[2]:=Clear["Global`*"];
Total[ParallelTable[
   If[TrueQ[
     Product[Part[IntegerDigits[n], k], {k, 1, 
         1 + Floor[Log10[n]]}] == 48 && 
      MemberQ[IntegerDigits[n], 1] == False], n, Nothing], {n, 1, 
    10^7}] //. {} -> Nothing]

Out[2]=179141

A: There isn't a much better way, no. There are better ways for calculating the number of ordered factorizations but those will include factorizations with factors that aren't single digits.
If $a_n = $ the number of ordered factorizations of $n$ then we have two identities:

*

*$a_n = \sum a_d$ over all factors $d$ of $n$ not including $d = n$.

*$a_{p^k} = 2^{k - 1}$ when $p$ is a prime.

See http://oeis.org/A074206 for references/additional formulae/etc.
So using this we have
\begin{align}
a_{48} &= a_1 + a_2 + a_3 + a_4 + a_6 + a_8 + a_{12} + a_{16} + a_{24} \\
&= 1 + 1 + 1 + 2 + a_6 + 4 + a_{12} + 8 + a_{24} \\
&= 17 + a_6 + a_{12} + a_{24} \\
&= 17 + (a_1 + a_2 + a_3) + ([a_1 + a_2 + a_3] + a_4 + a_6) \\
&\quad\qquad+ ([a_1 + a_2 + a_3 + a_4 + a_6] + a_8 + a_{12}) \\
&= 17 + 3 + (3 + 2 + 3) + (8 + 4 + 8) \\
&= 48
\end{align}
So there are 48 total ordered factorizations. But we know this is overcounting because it includes factorizations which are permutations of $2,24$ and $2,2,12$ and $4,12$ and $3,16$. So that's an additional $2 + 3 + 2 + 2 = 10$.
So there are $48 - 10 = 38$ ordered factorizations where no factor is larger than $9$.
Was this easier than calculating the number $38$ directly? Probably not. But hopefully it's still useful to someone.
