Write a shell script to find the "reverse order" of a number“

In an arrangement , If the front and back positions of a logarithm are in reverse order of magnitude , That is, the number in front is greater than the number in the back , Then they are called a reverse order . The total number in reverse order in an arrangement is called the number in reverse order of the arrangement . in other words , about n A different element , First, there is a standard order between the elements （ for example n individual Different natural numbers , The standard order can be set from small to large ）, So here it is n In any arrangement of elements , When the actual order of two elements is different from the standard order , That's all 1 In reverse order . The total number of all the inverses in an arrangement is called the reverse number of the permutation .
Such as 2431 in ,21,43,41,31 It's in reverse order , The reverse number is 4.

#!/bin/bash
echo -n  Please enter a number ：
read n
count=0 # Used to record the number in reverse order 
while [ $n -gt 0 ]
do
n1=`expr $n % 10` # Get the number of bits 
n2=`expr $n / 10` # Get other numbers except bits 
n=$n2
# One by one, it is better than the past 
while [ $n2 -gt 0 ]
do
n3=`expr $n2 % 10`
n2=`expr $n2 / 10`
if [ $n1 -lt $n3 ]
then
count=`expr $count + 1`
fi
done
done
echo  The reverse order of this number is ：$count