I was hoping that someone could give me some direction with how to solve this probem.

The only info in this problem is
word has N digits[only digits] and the numbers (output) should be in increasing order example 1469 or 468

The output should give all combinations in the increasing order

Example :
Enter length of number : 3

This output should give all numbers without recursion in increasing order for numbers between [0-9].


I believe you need to use recursive functions.

Start at the right, increment it by 1, then scan to the left to see what else can be incremented.

//This code will I guess increment the last digit example

The next number I guess would be to 023,024....
Can someone guide me as to how or where can I  make this change in the code.
void PrintNumbers (int length)
  int a[] ={0,1,2,3,4,5,6,7,8,9};
  char result [length + 1];
  int i;

  for (i=0;i<length;i++)
  result[i]= a[i];
 //Main loop begins here
  while (1) {
   printf (  "%d"\n",result);
  for (i=length-1 ; i>=-1;i--)
    if (i==-1) return;
    if (a[i] == result [i]) {
    resullt [i] == a[i+1];

Obviously recursion would enable a simpler/cleaner approach, if it were allowed.

There is no mention of need for speed/efficiency in the problem statement, so it may be perfectly valid to just use a dirty brute-force method:

#include <iostream>
#include <iomanip>
#include <sstream>
using namespace std;

bool numberValid(int v, int len){
   int oldd = 10;
   for(int i = 0; i < len; i++){
      int d = v % 10;
      v /= 10;

      if(d >= oldd)
         return false;     // non-increasing progression
      oldd = d;
   return true;      // valid
   int len;
   cout << "Enter len: ";
   cin >> len;
   int maxVal = (int[]){9, 89, 789, 6789, 56789, 456789, 3456789, 23456789, 123456789, 123456789}[len-1];

   for(int i = 0; i <= maxVal; i++){
      if(numberValid(i, len)){
         cout << setw(len) << setfill('0') << i << endl;

Start at the right, increment it by 1, then scan to the left to see what else can be incremented.

You would only scan to the left if a more-right digit couldn't be incremented - in which case you would try to increment a digit to the left. When you find a digit that can be incremented, you would increment it then set all digits to the right of it equal to 1 more than the digit to their left.

Apparently English is my first language, but good luck understanding what I've just written; I'm sure it made sense when I wrote it.

dougy83 thank you very much. That was a perfect !!!!