Sure, let's solve the problem step-by-step.
We are given the combination [tex]\({}^{3n}C_2 = 15\)[/tex]. The formula for combinations is given by [tex]\({}^nC_r = \frac{n!}{r!(n-r)!}\)[/tex].
Here, [tex]\(n = 3n\)[/tex] and [tex]\(r = 2\)[/tex]. Plugging in these values, we get:
[tex]\[
{}^{3n}C_2 = \frac{(3n)!}{2!(3n-2)!} = 15
\][/tex]
First, let's break this down. Using the factorial properties, we can write:
[tex]\[
\frac{(3n)!}{(3n-2)!} = (3n)(3n-1)(3n-2)!
\][/tex]
Thus, the equation simplifies to:
[tex]\[
{}^{3n}C_2 = \frac{(3n)(3n-1)(3n-2)!}{2!(3n-2)!} = \frac{(3n)(3n-1)}{2} = 15
\][/tex]
We can now eliminate the factorial terms [tex]\((3n-2)!\)[/tex] which appear in both the numerator and the denominator:
[tex]\[
\frac{(3n)(3n-1)}{2} = 15
\][/tex]
To eliminate the fraction, multiply both sides by 2:
[tex]\[
(3n)(3n-1) = 30
\][/tex]
Expanding the left-hand side:
[tex]\[
3n \cdot 3n - 3n = 9n^2 - 3n
\][/tex]
So the equation becomes:
[tex]\[
9n^2 - 3n = 30
\][/tex]
Subtract 30 from both sides to set the equation to 0:
[tex]\[
9n^2 - 3n - 30 = 0
\][/tex]
This is a quadratic equation in the standard form [tex]\(ax^2 + bx + c = 0\)[/tex], where [tex]\(a = 9\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -30\)[/tex].
To solve this quadratic equation, we can use the quadratic formula [tex]\(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[
n = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 9 \cdot (-30)}}{2 \cdot 9}
\][/tex]
Simplify inside the square root:
[tex]\[
n = \frac{3 \pm \sqrt{9 + 1080}}{18}
\][/tex]
[tex]\[
n = \frac{3 \pm \sqrt{1089}}{18}
\][/tex]
[tex]\[
\sqrt{1089} = 33, \text{ so:}
\][/tex]
[tex]\[
n = \frac{3 \pm 33}{18}
\][/tex]
This gives us two values for [tex]\(n\)[/tex]:
[tex]\[
n = \frac{3 + 33}{18} = \frac{36}{18} = 2
\][/tex]
[tex]\[
n = \frac{3 - 33}{18} = \frac{-30}{18} = -\frac{5}{3}
\][/tex]
Thus, we have two solutions: [tex]\( n = 2 \)[/tex] and [tex]\(n = -\frac{5}{3}\)[/tex]. However, since [tex]\(n\)[/tex] is a variable that represents a count (combinations and factorials only make sense for non-negative integers), we discard the negative fraction solution.
Therefore, the required value of [tex]\(n\)[/tex] is:
[tex]\[
n = 2
\][/tex]
