Test question: Number of possible combinations of numbers

Question

$3$ 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

Answer 1

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.

Answer 2

Let the number be of the form $2^{a_1}5^{b_1}{101}^{c_1}\ ,\ 2^{a_2}5^{b_2}{101}^{c_2}\ and\ 2^{a_3}5^{b_3}{101}^{c_3}$

So $a_1\ +\ a_2\ +\ a_3$ = 2(such that all three are whole numbers which gives 6 cases)

And b's and c's give 3 cases each

As a's , b's and c's are independent of each other , therefore total number of cases $6\times3\times3$= 54

Answer 3

You can think of it as giving out the prime factors. Each prime is independent, so you multiply the number of choices. For $5$ and $101$ you have three choices of who to give it to. For the $2$s, there are ${4 \choose 2}=6$ ways to pick three ordered numbers (including zero) that add up to $2$, so there are $3^2\cdot 6=54$ ways overall. You can get the $4 \choose 2$ by stars and bars.

Answer 4

Apply stars-and-bars and the rule of product.

Decide how to distribute the $2$ factors of two in $\binom{2+3-1}{3-1} = 6$ ways.

Decide how to distribute the $1$ factor of five in $\binom{1+3-1}{3-1}=3$ ways.

Decide how to distribute the $1$ factor one-hundred-and-one in $\binom{1+3-1}{3-1}=3$ ways.

This gives a total of $6\times 3\times 3$ total possible arrangements.

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This can be generalized. The number of ways of distributing factors of a number whose prime factorization is $p_1^{c_1}p_2^{c_2}\cdots p_n^{c_n}$ among $k$ people would be

$$\prod\limits_{i=1}^n \binom{c_i+k-1}{k-1}$$