C program to print all the possible permutations of given digits

Write a C program to print all the possible permutations of given digits.
Permutations means possible way of rearranging in the group or set in the particular order.
Example:
Input:1, 2, 3
Output:1 2 3, 1 3 2, 2 1 3, 3 1 2, 2 3 1, 3 2 1
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

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#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
int lev=-1,n,val[50],a[50];
void main()
{
int i,j;
clrscr();
printf("Enter how many numbers?n");
scanf("%d",&n);
printf("nEnter %d numbers:nn",n);
for(i=0;i<n;i++)
{
val[i]=0;
j=i+1;
scanf("%dnn",&a[j]);
}
visit(0);
getch();
}
visit(int k)
{
int i;
val[k]=++lev;
if(lev==n)
{
for(i=0;i<n;i++)
printf("%2d",a[val[i]]);
printf(" ");
}
for(i=0;i<n;i++)
if(val[i]==0)
visit(i);
lev--;
val[k]=0;

}
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C Program to calculate the Combinations and Permutations.

C Program to calculate the Combination and Permutations. Combination means way of selecting a things or particular item from the group or sets. nCr=n!/r!(n-r)!. Permutations means possible way of rearranging in the group or set in the particular order. Read more about C Programming Language .

/***********************************************************
* You can use all the programs on www.c-program-example.com
* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact [email protected]
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/

#include <stdio.h>
#include <conio.h>

main()
{
int n , r, ncr( int , int);
long npr( int , int);
long double fact( int);
printf(" Enter value of n & r n");
scanf("%d %d",&n , &r);
if( n>= r)
{
printf( " %dC%d is %dn", n,r,ncr( n , r));
printf(" %dP%d is %ld", n,r,npr( n, r));
}
else
{
printf("WRONG INPUT?? enter the correct input");
}
}
long double fact( int p)
{
long double facts = 1;
int i;
for( i = 1; i<= p; i++)
facts = facts * i;
return( facts);
}

int ncr ( int n, int r)
{
return( fact( n) / (fact( r) * fact(n- r) ) ) ;
}

long npr( int n , int r)
{
return( fact( n) / fact( n- r));
}
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