Certainly! Let's find the values step-by-step for each part of the question.
### Part a: [tex]\(8!\)[/tex]
The factorial of a number [tex]\( n \)[/tex], denoted as [tex]\( n! \)[/tex], is the product of all positive integers less than or equal to [tex]\( n \)[/tex].
So, for [tex]\( 8! \)[/tex]:
[tex]\[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
Performing this multiplication,
[tex]\[ 8! = 40320 \][/tex]
Thus, [tex]\( 8! = 40320 \)[/tex].
### Part b: [tex]\(6!\)[/tex]
Now, calculating the factorial of [tex]\( 6 \)[/tex]:
[tex]\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
Performing this multiplication,
[tex]\[ 6! = 720 \][/tex]
So, [tex]\( 6! = 720 \)[/tex].
### Part c: [tex]\(\frac{7!}{4!}\)[/tex]
For this part, we need to find the value of the division of two factorials, [tex]\( 7! \)[/tex] and [tex]\( 4! \)[/tex].
First, let's write down the factorials:
[tex]\[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
[tex]\[ 4! = 4 \times 3 \times 2 \times 1 \][/tex]
So, [tex]\(\frac{7!}{4!}\)[/tex] is:
[tex]\[ \frac{7!}{4!} = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} \][/tex]
Notice that the [tex]\( 4!, \)[/tex] or [tex]\( 4 \times 3 \times 2 \times 1, \)[/tex] cancels each other out in the numerator and denominator,
[tex]\[ \frac{7!}{4!} = 7 \times 6 \times 5 \][/tex]
Performing this multiplication,
[tex]\[ 7 \times 6 \times 5 = 210.0 \][/tex]
So, [tex]\(\frac{7!}{4!} = 210.0 \)[/tex].
### Summary
- [tex]\( 8! = 40320 \)[/tex]
- [tex]\( 6! = 720 \)[/tex]
- [tex]\(\frac{7!}{4!} = 210.0 \)[/tex]
These are the detailed steps and final values for the given parts of the problem.
