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Chapter 9: Problem 79

Solve for \(n\). $$_{n} P_{4}=10 \cdot_{n-1} P_{3}$$

Short Answer

Expert verified
The value of \(n\) that satisfies the given equation is \(n = 10\).

Step by step solution

01

Replace permutations with their factorial equivalents

Let's replace the permutations on both sides of the equation with their factorial equivalents. Now, our equation becomes: \[\frac{n!}{(n-4)!} = 10 \cdot \frac{(n-1)!}{(n-4)!}\]. Note that the (n-4)! gets canceled on both sides.
02

Simplify the equation

After canceling (n-4)! on both sides, the equation simplifies to: \[n! = 10 \cdot (n-1)!\].
03

Use the property of factorial

Expanding n! on the left side of the equation using the property of factorial (n! = n*(n-1)!), we obtain: \[n \cdot (n-1)! = 10 \cdot (n-1)!\].
04

Solve for n

By further simplifying the equation, we get: \(n = 10\).

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