Problem 78 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve for \(n\). $$_{n} P_{5}=18 \cdot_{n-2} P_{4}$$

Step-by-Step Solution

Answer

The solution for \(n\) is 10.

Step 1: Convert to Factorial Notation

To convert the permutation notation to factorial notation, write the permutation function \(_{n} P_{m}\) as a ratio of factorials \( \frac{n!}{(n-m)!}\). So, the equation \(_{n} P_{5}=18 \cdot_{n-2} P_{4}\) becomes \( \frac{n!}{(n-5)!} = 18 \cdot \frac{(n-2)!}{[(n-2)-4]!}\)

Step 2: Simplify Equation

Now, simplify the equation, \( \frac{n!}{(n-5)!} = 18 \cdot \frac{(n-2)!}{(n-6)!}\), becomes: \( \frac{n \cdot (n-1) \cdot (n-2) \cdot (n-3) \cdot (n-4)}{ 1 } = 18 \cdot (n-2) \cdot (n-3) \cdot (n-4) \cdot (n-5) \). This simplifies even further to \( n(n-1) = 18(n-5)\).

Step 3: Solve for n

The equation above can be simplified to get a quadratic in the form \( ax^{2} + bx + c = 0\): \(n^2 - n = 18n - 90\).Re-order and simplify to: \(n^2 - 19n + 90 = 0\). This quadratic equation can be factored to: \((n-10)(n-9) = 0\). Therefore, \(n = 10\) or \(n = 9\), but the solution is \(n = 10\) because a permutation cannot have negative values or values smaller than the number of ordered outcomes.