Calculating possible combinations

[Amidda1217]

I know this can probably be done via factorial. But it has been a long time and I have forgotten how to do it.

If you have 2 variables x and y of two sets. So x1 x2 y 1 y2 and you want to test all possible combinations of > = <. How do you calculate the possible number of combinations?

x1 > y1 and
x2 > y2

x1 < y1 and
x2 < y2

X1 >y1 and
x2 < y1

x1 < y1and
x2 >y1

x1 = y1and
x2 = y2

x1 > y1and
x2 = y2

and so on...

[cilpolir]

I would say X1 can be <,= or > to Y1, sot that are 3 combinations, X1 can also be that to y2 so that is 3*3 combinations, repeating this for x2 gives 3^4 combinations
edit:
I wrote a code in c to write down al combinations and count them, here's the code:

Code: Select all

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

int main()
{
    char signs[3];
    char sign1,sign2,sign3,sign4;
    int a,b,c,d;
    int combination;
    FILE *fp;

    fp=fopen("test.txt","w");

    signs[0]='>';
    signs[1]='=';
    signs[2]='<';

    combination=0;

    for(a=0;a<3;a++)
    {
        for(b=0;b<3;b++)
        {
            for(c=0;c<3;c++)
            {
                for(d=0;d<3;d++)
                {
                    sign1=signs[a];
                    sign2=signs[b];
                    sign3=signs[c];
                    sign4=signs[d];
                    fprintf (fp, "x1 %c y1\nx1 %c y2\nx2 %c y1\nx2 %c y2\n\n",sign1,sign2,sign3,sign4);
                    combination++;
                }
            }
        }
    }
    fprintf (fp, "there is a total of %d combinations",combination);
    fclose (fp);
    return 0;

}

[msbachman]

This sounds like a homework question.

@ the OP: what are YOUR ideas about how to calculate the possible combinations?

Give the forum something to show you've done some work on this issue on your own. That said, clipolir's response was very helpful.

@ Clipolir:

If you like solving problems like that, check out freelancer.com. That way you can help people and make some money, instead of helping people and having them take advantage of you. I think the latter is in play here.

-- Fri Jul 30, 2010 7:42 pm --

Case in point:

I know this can probably be done via factorial. But it has been a long time and I have forgotten how to do it.

So, what the OP is basically saying is this:

...I've forgotten how to do it...and since my time is worth so much more than everyone else's, I'll post it onto the HTS forums for someone else to bother with it.

I can't be bothered to put forth effort to solve my own problems. Someone, solve this for me and I'll get to it when it's convenient for me.

I'm not saying that the OP isn't worth the help, it's just that at this point there's not much to evidence that he/she worked hard at solving it independently before asking us. If "factorials" were an answer to this issue (I'd like to see how they are, btw), then they don't need our help. They already have the answer. The work is up to them.