P(9,3)/6!= ?  I tried using the n!/(n-r)! for the P and got 504 for the overall answer, but it was wrong.

A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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P(9,3)/6!= ?
 I tried using the n!/(n-r)! for the P and got 504 for the overall answer, but it was wrong. 

 

Expert Solution
Step 1

For the numerator we will sue the formula:

P(n,r)=n!(n-r)!

 

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