If \(^{3x}C_{2} = 15\), find the value of x?

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.