A probability problem by Karan Siwach

We can factorization $120$ into $2^3 \times 3 \times 5$ Which told us $a$ have $4 \times 2 \times 2 = 16$ possibilities.

Let consider three variables $p, q, r$ which range are $[0,3], [0,1], [0,1]$ respectively.

So we want to count the total number of all of the possibilities of $p, q, r$ such that

$xyz = 2^p \times 3^q \times 5^r$

Let $x = 2^{p_x} 3^{q_x} 5^{r_x}$ , $y = 2^{p_y} 3^{q_y} 5^{r_y}$ , $z = 2^{p_z} 3^{q_z} 5^{r_z}$ for integers variables inside that.

After that we have equations:

$p_x + p_y + p_z = p$

$q_x + q_y + q_z = q$

$r_x + r_y + r_z = r$

So for each tuple $(p, q, r)$ , the number of ordered tuple $(x, y, z)$ is $C_2^{p+2} \times C_2^{q+2} \times C_2^{r+2}$

Sum up all of that, the result is equivalent to

$(C_2^2 + C_2^3 + C_2^4 + C_2^5)(C_2^2 + C_2^3)(C_2^2 + C_2^3) = \boxed{320}$

120 = 2^(3) x 3 x 5

Therefore, total number of factors of 120 = (3+1) x (1+1) x (1+1) = 16

But a can be any of the 16 factors.

Now we use combination to assess how the factors are distributed if any a among the 16 factors is chosen.

Observe that there are five prime factor: 2,2,2,3 and 5. These can take any of the 3 variables x, y or z.

So, (5+1)C3 = 20 ; 1 has been added to account for 1 itself. For example 8 has 4 factors: 1,2,4 and 8.

Therefore, K = 20 x 16 = 320