Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 11 - Counting Methods and Probability Theory - 11.2 Permutations - Concept and Vocabulary Check - Page 700: 6


The statement is true.

Work Step by Step

Permutation problems are arrangement problems which can be solved by observing a successive list of questions: 1. In how many ways can we choose the 1st object? Answer: $n$. 2. In how many ways can we choose the 2nd object? Answer: $n-1$. ... r. In how many ways can we choose the rth object? Answer: $n-r+1$. The total number of ways is, using FCP, $n(n-1)n-2)\cdot...\cdot(n-r+1)$, The formula for ${}_{n}P_{r}$, (the number of ordered sequences taking r objects from n, each not being used more than once.), leads to the same expression : ${}_{n}P_{r}=\displaystyle \frac{n!}{(n-r)!}=$ $=\displaystyle \frac{n(n-1)\cdot...\cdot ...(n-r+1)(n-r)!}{(n-r)!}$ ... after reducing $(n-r)!$, ... $= n(n-1)(n-2)...(n-r+1)$ The statement is true.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.