$3$ distinct people are given a positive integer. How many ways are
there to have the $3$ numbers form a product of $2020$?
Here is what I have tried:
$2020$ = $2$ * $2$ * $5$ * $101$
So I must give numbers to the $3$ people in many different combinations.
E.g: 4,5,101 or 2,10,101 etc etc.
The problem is: The number $1$ is also a valid number to give each person.
Is there a neat way to find out the number of combinations to get a product of $2020$?
Note: Giving 1,1,2020 is different from 1,2020,1
Best Answer
There are 3 ways to distribute the factor of 101. There are 3 ways to distribute the factor of 5. There are 6 ways to distribute the two factors of 2 (either one person gets both factors in 3 ways or two people each get a factor of two in 3 ways).
By the product principle, there are $3\cdot 3\cdot 6 = 54$ different combinations of factors.