Number of unique digits numbers 1-5324
=   Number of unique digits numbers 1-9
  + Number of unique digits numbers 10-99
  + Number of unique digits numbers 100-999
  + Number of unique digits numbers 1000-5324

f(n) = Number of unique digits numbers with length n.
f(1) = 9 (1-9)
f(2) = 9*9 (1-9 * 0-9 (excluding first digit))
f(3) = 9*9*8 (1-9 * 0-9 (excluding first digit) * 0-9 (excluding first 2 digits))
f(4) = 9*9*8*7

Number of unique digits numbers 1000-5324
=   Number of unique digits numbers 1000-4999
  + Number of unique digits numbers 5000-5299
  + Number of unique digits numbers 5300-5319
  + Number of unique digits numbers 5320-5324

N = 5324

If N[0] = 1, there are 9*8*7 possibilities for the other digits
If N[0] = 2, there are 9*8*7 possibilities for the other digits
If N[0] = 3, there are 9*8*7 possibilities for the other digits
If N[0] = 4, there are 9*8*7 possibilities for the other digits
If N[0] = 5
  If N[1] = 0, there are 8*7 possibilities for the other digits
  If N[1] = 1, there are 8*7 possibilities for the other digits
  If N[1] = 2, there are 8*7 possibilities for the other digits
  If N[1] = 3
    If N[2] = 0, there are 7 possibilities for the other digits
    If N[2] = 1, there are 7 possibilities for the other digits
    If N[2] = 2
      If N[3] = 0, there is 1 possibility (no other digits)
      If N[3] = 1, there is 1 possibility (no other digits)
      If N[3] = 2, there is 1 possibility (no other digits)
      If N[3] = 3, there is 1 possibility (no other digits)

uniques += (N[0]-1)*9!/(9-N.length+1)!
for (int i = 1:N.length)
  uniques += N[i]*(9-i)!/(9-N.length+1)!

// don't forget N
if (hasUniqueDigits(N))
  uniques += 1

binary used[10]
uniques += (N[0]-1)*9!/(9-N.length+1)!
used[N[0]] = 1
for (int i = 1:N.length)
  uniques += (N[i]-sum(used 0 to N[i]))*(9-i)!/(9-N.length+1)!
  if (used[N[i]] == 1)
    break
  used[N[i]] = 1

// still need to remember N
if (hasUniqueDigits(N))
  uniques += 1

Prelude> :m +Data.List
Data.List> length [a | a <- [1..5324], length (show a) == length (nub $ show a)]
2939

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
 public static void main(String[] args) {

         int rem;
        Scanner in=new Scanner(System.in);
         int num=in.nextInt();
    int length = (int)(Math.log10(num)+1);//This one is to find the length of the number i.e number of digits of a number
    // System.out.println(length);
    int arr[]=new int[length]; //Array to store the individual numbers of a digit for example 123 then we will store 1,2,3 in the array

    int count=0;
     int i=0;

     while(num>0)           //Logic to store the digits in array
    { rem=num%10;
        arr[i++]=rem;
        num=num/10;
    }
    for( i=0;i<length;i++)          //Logic to find the duplicate numbers
    {
        for(int j=i+1;j<length;j++)
        {
            if(arr[i]==arr[j])
            {
                count++;
            }
        }
    }
     //Finally total number of digits minus duplicates gives the output
     System.out.println(length-count);
   }
}