The formula for finding the number of combinations of r items out of n distinct items is given by the formula:
$$^{n}C_{r} = \frac{n!}{r!(n-r)!}$$
Where n! is the factorial of n, that is the product of all positive integers from 1 to n.
In this question, we are given that $$^{3x}C_{2} = 15$$
Substituting the given values into the formula above, we have:
$$^{3x}C_{2} = \frac{(3x)!}{2!(3x-2)!} = 15$$
Simplifying this equation, we have:
$$\frac{(3x)(3x-1)(3x-2)!}{2!} = 15$$
Multiplying both sides by 2, we get:
$$(3x)(3x-1)(3x-2)! = 30$$
We can observe that 3x-2! is the factorial of (3x-2) which is an integer, and 3x-1 and 3x are consecutive integers. Therefore, we can re-write the equation above as:
$$(3x)(3x-1) = 10$$
Expanding the left-hand side, we have:
$$9x^2 - 3x = 10$$
Bringing all the terms to one side, we get:
$$9x^2 - 3x - 10 = 0$$
We can then factorize this quadratic equation as follows:
$$(3x - 5)(3x + 2) = 0$$
Using the zero-product property, we get:
$$3x - 5 = 0 \quad \text{or} \quad 3x + 2 = 0$$
Solving for x in each equation, we get:
$$x = \frac{5}{3} \quad \text{or} \quad x = -\frac{2}{3}$$
Since x represents the number of items in the set from which we are selecting combinations, it must be a positive integer. Therefore, the only valid solution is:
$$x = \frac{5}{3} = 1.67 \approx 2$$
Hence, the value of x is 2.