Question

Find the value of \(n\) if \[ {}^{n}P_{4} = 12 \times {}^{n}P_{2}. \]

Hint:

For permutation equations, convert each term using \[ {}^{n}P_{r}=\frac{n!}{(n-r)!} \] Then simplify factorial expressions carefully to form an algebraic equation in \(n\).

Answer 1

Concept: The permutation formula is \[ {}^{n}P_{r}=\frac{n!}{(n-r)!} \] Using this formula, we express both permutations and then solve the resulting equation.
Step 1: Write the permutations using factorial form. \[ {}^{n}P_{4}=\frac{n!}{(n-4)!} \] \[ {}^{n}P_{2}=\frac{n!}{(n-2)!} \] Substituting into the given equation: \[ \frac{n!}{(n-4)!}=12\times\frac{n!}{(n-2)!} \]
Step 2: Simplify the equation. Cancel \(n!\) from both sides: \[ \frac{1}{(n-4)!}= \frac{12}{(n-2)!} \] Since \[ (n-2)!=(n-2)(n-3)(n-4)! \] Substitute this into the equation: \[ \frac{1}{(n-4)!}=\frac{12}{(n-2)(n-3)(n-4)!} \] Cancel \((n-4)!\): \[ 1=\frac{12}{(n-2)(n-3)} \]
Step 3: Solve for \(n\). \[ (n-2)(n-3)=12 \] \[ n^2-5n+6=12 \] \[ n^2-5n-6=0 \] \[ (n-6)(n+1)=0 \] \[ n=6 \quad \text{or} \quad n=-1 \]
Step 4: Select the valid value. Since \(n\) represents the number of objects, it must be positive. \[ \boxed{n=6} \]