7! - 41 =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![**Problem: Evaluate the factorial expression.**
\[ 7! - 4! \]
---
**Solution:**
To evaluate the expression \(7! - 4!\), we need to calculate the factorial of each number and then subtract.
- \(7!\) (7 factorial) is the product of all positive integers from 1 to 7:
\[
7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040
\]
- \(4!\) (4 factorial) is the product of all positive integers from 1 to 4:
\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]
Now, subtract the two values:
\[
7! - 4! = 5040 - 24 = 5016
\]
Place the answer in the box.
\[ \boxed{5016} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F022ae44f-636d-4b99-a07b-5f1c14bbf99e%2Fa58a8965-090a-4339-81db-fa04292aad30%2Fekocf54_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem: Evaluate the factorial expression.**
\[ 7! - 4! \]
---
**Solution:**
To evaluate the expression \(7! - 4!\), we need to calculate the factorial of each number and then subtract.
- \(7!\) (7 factorial) is the product of all positive integers from 1 to 7:
\[
7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040
\]
- \(4!\) (4 factorial) is the product of all positive integers from 1 to 4:
\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]
Now, subtract the two values:
\[
7! - 4! = 5040 - 24 = 5016
\]
Place the answer in the box.
\[ \boxed{5016} \]
Expert Solution
Step 1
We have to evaluate the following factorial expression:
Note that
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