This article will depict a fast and simple calculation that gives the full stage for a characteristic number. The calculation will be actualized by C++.

A full change is rundown of all variety forgiven things (typically numbers). For instance, the full stage of 3 components are:

```
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
```

As we can undoubtedly compute, the absolute number of emphasis is n! (factorial). How would we get this number?

The most straightforward path is to think about the number of various things we can decide for each position.

For instance, for the primary spot, we have n decisions, and we have n-1 decisions for the subsequent spot, and so forth.

Consequently we have n(n-1)(n-2)*… 1 that gives an all outnumber n!.

So how to imitate this for the entire procedure? In the first place, we can characterize a recursive

the capacity that finishes printing one arrangement by checking on the off chance that we have arrived at the end.

For every emphasis, we can just swap the present thing with the remainder of the things and attempt the following position recursive.

As we utilize a worldwide cluster variable nums to keep the things, we have to swap the things back after every recursion call.

The source code is truly straight forward. Open Visual Studio, pick the ‘Comfort Application’ from ‘C++’ language ventures.

What’s more, glue the accompanying source code into Visual Studio.

**Program**

```
#include <iostream>
using namespace std;
// helper dynamic array
int *nums;
// exchange two integers
inline void swap(int &i, int &j)
{
int t = i;
i = j;
j = t;
}
// full permutation
void
perm(int n, int i)
{
if (i == n – 1) //
check if reach the end of iteration
{
// print out
the iteration
for (int j =
0; j < n; j ++)
{
cout
<< nums[j];
}
cout <<
endl;
}
else
{
for (int j =
i; j < n; j ++)
{
swap(nums[i],
nums[j]); // swap
perm(n,
i + 1); // recursive
swap(nums[i],
nums[j]); // swap it back
}
}
}
// main entry
int
main()
{
int n;
cin >> n;
nums = new int[n]; //
declare a dynamic array
for (int i = 0; i
< n; i ++)
{
// and fill it
with numbers as 1, 2, 3, … n
nums[i] = i +
1;
}
perm(n, 0);
cin >> n; //
pause
return 0;
}
```

Press F5 to run the venture, put a number, for example, 4, and the program will give the full stage of 4. The keep going cin >> n is the C++ simple approach to delay the screen after the program ends.

**Output**