Certainly! To solve the expression [tex]\(1! \cdot (n-1)!\)[/tex] and to understand how we get to the final result, follow these steps:
1. Understand Factorial: The factorial of a number [tex]\( n! \)[/tex] is the product of all positive integers less than or equal to [tex]\( n \)[/tex]. So for example:
[tex]\[
1! = 1
\][/tex]
Factorials grow very quickly, and it's important to understand the basic definition before applying it.
2. Evaluate [tex]\( 1! \)[/tex]:
- By definition, [tex]\( 1! = 1 \)[/tex].
3. Evaluate [tex]\( (n-1)! \)[/tex]:
- Let's take [tex]\( n = 5 \)[/tex] as the given value.
- Then [tex]\( (n-1) = 4 \)[/tex].
- So, [tex]\( 4! = 4 \times 3 \times 2 \times 1 \)[/tex].
4. Compute [tex]\( 4! \)[/tex] (continuation of the factorial computation):
- [tex]\( 4! = 4 \times 3 \times 2 \times 1 = 24 \)[/tex].
5. Multiply Both Results:
- Using the values we calculated:
[tex]\[
1! = 1
\][/tex]
[tex]\[
4! = 24
\][/tex]
- Many [tex]\[ 1 \cdot 24 = 24 \][/tex].
Therefore, for [tex]\( n = 5 \)[/tex], the expression [tex]\( 1!(n-1)! \)[/tex] evaluates to [tex]\( 1 \cdot 24 = 24 \)[/tex].
The answer as a pair of factorial results would be [tex]\((1, 24)\)[/tex].
