To solve the given problem, we use the formula for permutations:

$^nP_r = \frac{n!}{(n-r)!}$

The given equation is:

$^nP_4 = 12 \cdot ^nP_2$

Substituting the permutation formula:

$\frac{n!}{(n-4)!} = 12 \cdot \frac{n!}{(n-2)!}$

Simplify by canceling $n!$ on both sides:

$\frac{1}{(n-4)!} = 12 \cdot \frac{1}{(n-2)(n-3)(n-4)!}$

Simplify further by canceling $(n-4)!$:

$1 = \frac{12}{(n-2)(n-3)}$

Multiply through by $(n-2)(n-3)$:

$(n-2)(n-3) = 12$

Expand and simplify:

$n^2 - 5n + 6 = 12$

$n^2 - 5n - 6 = 0$

Factorize:

$(n-6)(n+1) = 0$

So:

$n = 6 \quad \text{or} \quad n = -1$

Since $n$ must be a positive integer, $n = 6$.

Final Answer:

$n = 6$

Would you like a detailed breakdown of any step?

Here are 5 related questions for deeper exploration:

1. What is the difference between permutations and combinations?
2. How do factorials simplify expressions involving permutations?
3. Can this method be used to solve any permutation-based equation?
4. What happens if the values of $r$ in the permutations differ significantly?
5. How do negative results for $n$ affect practical problems?

Tip: Always ensure that factorial terms in permutations are simplified to avoid overly complex equations.