Printing all permutations of n numbers

Print all n! permutations of the number 1,2,3,...,n.
Example: Input: 3
Output: 1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1

Following is my approach. My program is not working for inputs greater than 3. I understand the logic why it is not working , but I am unable to translate that logic into a code block to overcome that issue.

#include <stdio.h>
int permute(int n)
{
   int a[n];
   int i,j,k,store;
   for(i=0;i<n;i++)
    a[i]=i+1;
   for(i=1;i<=n;i++)
   {
      for(j=0;j<n-1;j++)
      {
         store=a[j+1];
         a[j+1]=a[j];
         a[j]=store;
         for(k=0;k<n;k++)
         printf("%d ",a[k]);
         printf("\n");
      }
      
   }

}
int main()
{
   int n;
   scanf("%d",&n);
   permute(n);
   return 0;
}

Following is the output for n as 4:

We can clearly see that some permutation are missing, and I know exactly the fault in my code. But I am unable to fix it.( I am a beginner , hence I don't know much advanced C libraries or functions)

One solution consists in calling the function recursively: you set the first number (n possible choices), then call the function for a size n-1.

Output, for n=4

1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
1 4 3 2
1 4 2 3
2 1 3 4
2 1 4 3
2 3 1 4
2 3 4 1
2 4 3 1
2 4 1 3
3 2 1 4
3 2 4 1
3 1 2 4
3 1 4 2
3 4 1 2
3 4 2 1
4 2 3 1
4 2 1 3
4 3 2 1
4 3 1 2
4 1 3 2
4 1 2 3

#include <stdio.h>
#include <stdlib.h>

void swap (int *i, int *j) {
    int temp = *i;
    *i = *j;
    *j = temp;
}

void permute(int index, int* arr, int n) {
   if (index == n-1) {
       for (int k = 0; k < n; ++k) {
            printf ("%d ", arr[k]);
       }
       printf ("\n");
       return;
   }
   for (int i = index; i < n; i++) {
       swap (arr + index, arr + i);
       permute (index+1, arr, n);
       swap (arr + i, arr + index);
   }
   return;
}

int main()
{
   int n;
   if (scanf("%d",&n) != 1) exit (1);
   int arr[n];
   for (int i = 0; i < n; ++i) arr[i] = i+1;
   permute(0, arr, n);
   return 0;
}