Find the number of permutations. 7 objects taken 4 at a time There are permutations.

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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**Problem Statement:**

*Find the number of permutations.*

**Details:**

*7 objects taken 4 at a time*

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**Answer Field:**

*There are [ ] permutations.*

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**Explanation:**

This problem requires finding the number of permutations of 7 distinct objects taken 4 at a time. To find the number of permutations, use the formula for permutations:

\[ P(n, r) = \frac{n!}{(n-r)!} \]

In this case, \( n = 7 \) and \( r = 4 \). 

Calculate as follows:

\[ P(7, 4) = \frac{7!}{(7-4)!} = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} = 7 \times 6 \times 5 \times 4 = 840 \]

Therefore, there are 840 permutations.
Transcribed Image Text:**Problem Statement:** *Find the number of permutations.* **Details:** *7 objects taken 4 at a time* --- **Answer Field:** *There are [ ] permutations.* --- **Explanation:** This problem requires finding the number of permutations of 7 distinct objects taken 4 at a time. To find the number of permutations, use the formula for permutations: \[ P(n, r) = \frac{n!}{(n-r)!} \] In this case, \( n = 7 \) and \( r = 4 \). Calculate as follows: \[ P(7, 4) = \frac{7!}{(7-4)!} = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} = 7 \times 6 \times 5 \times 4 = 840 \] Therefore, there are 840 permutations.
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