1. ## Combinations with factorial

Would somebody help me with this.

Let's take 4! (4 times 3 times 2 times 1 time = 24). This is easy to do but I want to have all 24 combinations on the screen.

Thanks

2. Code:
```for (x=2; x>0; x--)
{ for (y=2; y>0; y--)
{ if(y==x) continue;
cout<<x<<","<<y<<endl;
}
}```
There is a little code that will do it for 2!. You can expand on that for 4!.

3. You are creating *permutations*.
Here is a function for creating all permutations in order (there is one in STL too called next_permutation using templates):

Code:
```#include <iostream>
void swap(int &n1,int &n2) {
int temp=n1;
n1=n2;
n2=temp;
}
bool next_permutation(int *table, int length) {
int key=length-1, i=length-1;
while(key>0 && table[key]<=table[key-1])
key--;
if(key==0)return false;
key--;
while(table[key]>=table[i])
i--;
swap(table[key],table[i]);
for(i=length-1,key++;key<i;key++,i--)
swap(table[key],table[i]);
return true;
}
int main () {
int i, sequence[10]={1,2,3,4,5,6,7,8,9,10};
do {
for(i=0;i<10;i++)
std::cout<<sequence[i]<<" ";
std::cout<<std::endl;
} while(next_permutation(sequence,10));
return 0;
}```

4. ## Factorial

Thanks for the solutions. Is it possible to get a solution so I don't have to specify in advance the number of elements in the array.
Fero

5. I''m just guessing 'cuz I didn't look carefully (I'm probably wrong, but whatever), but looking at golfinguy's post, try this:

Code:
```void showStuff(int n)
{

for (x=n; x>0; x--)
{ for (y=n; y>0; y--)
{ if(y==x) continue;
cout<<x<<","<<y<<endl;
}
}

}```
again, this is a wild guess based on a quick glance at golfinguy's post...