Combinatorics: possible combinations of integers with set product

Question

I'm currently struggling with combinatorics, specifically with the problem:

How many are the triplets of positive integers (x,y,z) such that xyz=84?

And others of the same type. Does it have to do with congruences?

Answer 1

Any factor of $84 = 2^2 \cdot 3 \cdot 7$ has the form $2^{\alpha}3^{\beta}7^{\gamma}$. If the three factors are $x = 2^{\alpha_1}3^{\beta_1}7^{\gamma_1}$, $y = 2^{\alpha_2}3^{\beta_2}7^{\gamma_2}$, and $z = 2^{\alpha_3}3^{\beta_3}7^{\gamma_3}$, then \begin{align*} \alpha_1 + \alpha_2 + \alpha_3 & = 2 \tag{1}\\ \beta_1 + \beta_2 + \beta_3 & = 1 \tag{2}\\ \gamma_1 + \gamma_2 + \gamma_3 & = 1 \tag{3} \end{align*} Equations 1, 2, and 3 are equations in the nonnegative integers. Clearly, equations 2 and 3 each have three solutions, depending on which of the three variables is equal to 1. As for equation 1, a particular solution corresponds to the placement of two addition signs in a row of two ones. For instance, $$1 + + 1$$ corresponds to the solution $\alpha_1 = 1$, $\alpha_2 = 0$, $\alpha_3 = 1$, while $$1 1 + +$$ corresponds to the solution $\alpha_1 = 2$, $\alpha_2 = \alpha_3 = 0$. Thus, the number of solutions of equation 1 is the number of ways we can place two additions signs in a row of two ones, which is $$\binom{4}{2} = 6$$ since we must choose which two of the four positions required for two ones and two addition signs will be filled with addition signs. Hence, the number of ordered triples $(x, y, z)$ such that $xyz = 84$ is $6 \cdot 3 \cdot 3 = 54$, as you found.

Answer 2

How many ways can 84 = 2×2×3×7 be factored into three positive integers?