All the 5 positive integers that the product of their digits is 600? How many 5-digit positive integers have the property that the product of their digits is 600?
I know that this problem is based on Combinatorics, but I can't think of the approach and calculation part. Any sort of help is appreciated.
I did upto the Prime Factorisation of 600, i.e. $2^3 \times 5^3\times 3$.
 A: You have the $6$ numbers ${2,2,2,3,5,5}$ in the prime factorization if $600$. Now, consider any $6$ digit number with these numbers, for instance-
$${5~5~3~2~2~2}~~~\text{where,}~5\times5\times3\times2\times2\times2=600.$$
Now, for a $5$ digit number one of the digits has to be removed. However, you need to keep the product of digits equal and deleting a digit would change the final product. The way to achieve that is to club two digits such that their product forms a single digit. To understand this better, look at this 'clubbing' in the number given above-
$$5\times5\times3\times2\times(2\times2)=600~~~\text{giving,}~5~5~3~2~4$$
since, $2\times2=4$, a single digit number.
Hence, simply note the possibilities where the clubbing results in a single digit product for the two numbers allowing us to replace two of them with that number. We only have-
$$2\times2=4~\text{and}~2\times3=6.$$
Therefore, the possible digits for the required number involve digits-

*

*$4,2,3,5,5$: total number of possible numbers with these digits $=5!/2!$

*$2,2,6,5,5$: total number of possible numbers with these digits $=5!/(2!2!)$
Now, note that the prime factorization of any number can easily be modified to include a $1$, since it makes no change in a product. We can as a result, perform clubbing in two ways since, we have $7$ numbers to choose from-
$$5\times5\times3\times(2\times2\times2)\times1=600~~~\text{giving,}~5~5~3~8~1$$
$$\text{or,}~5\times5\times(3\times2)\times(2\times2)\times1=600~~~\text{giving,}~5~5~6~4~1$$
Total number of possible numbers with these digits is equal to $5!/2!$ for each case.
Sum up number of permutations possible casewise to get the final answer.
