Chapter 11.6, Problem 41E

To determine

To calculate: how many times greater is $n!$ than $\left(n-1\right)!$ .

Answer to Problem 41E

The $n!$ is $n$ time more greater than $\left(n-1\right)!$ .

Explanation of Solution

Given :

$n!$ and $\left(n-1\right)!$ .

Formula used:

  $n!=n·\left(n-1\right)!$

Calculation : tofind that, it is clearly by the formula that when finding the $n!$ , it iswritten like

  $\begin{array}{l}n!=n·\left(n-1\right)!\\ n!=n·\left(n-1\right)!\text{ [expand fatorial]}\\ \left(n-1\right)!=\left(n-1\right)!\end{array}$ .

Hence,The $n!$ is $n$ time more greater than $\left(n-1\right)!$ .