How can I convert a combination function to permutation function?

Question

I'm working on a separate program which has a bunch of gui, so I made a simpler program in an attempt to ask for help.

I don't really know permutations and combinations that well.

Anyway, I have a piece of code here which solves this function:

F (n,r) = n!/(r!(n-r)!)

You can change the n and r in the code and it'll give you the answer for those two inputs.

public class peanuts
{

static int n,r;


public static void main (String[] args)
{

   n= 8;
    r=3;

  int m=1;

    if(n>>1 < r)
        r = n - r;
    for(int o=n+1-r;o<n+1;o++)
        m *= o;
    for(int o=r;o>0;o--)
        m /= o;
    Math.round(m);

      System.out.println ("C("+n+","+r+ ") is "+ m );
}}

I want to modify it slightly so it solves this permutation function:

F (n,r) = n!/(n-r)!)

Question for future question writing, why did you put all that non-code into code format? — munchschair, Dec 19 '13 at 16:12
I guess he did it for readability though you can use `blockquote` for such purpose — exexzian, Dec 19 '13 at 16:17

score 2 · Accepted Answer · answered Dec 19 '13 at 16:05

2

Remove the second for loop that divides by r!.

for(int o=r;o>0;o--)
  m /= o;

This divides m by r!. Remove this loop.

answered Dec 19 '13 at 16:05

Abhishek Bansal

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Remove this? `for(int o=r;o>0;o--) m /= o;` – munchschair Dec 19 '13 at 16:06

score 2 · Answer 2 · answered Dec 19 '13 at 16:07

Formula for calculating the permutation is:

F (n,r) = n!/(n-r)!

Formula for calculating the combination is:

F (n,r) = n!/(r!(n-r)!)

So you just need to remove the second for loop in which you are dividing by factorial r ie, remove this:

for(int o=r;o>0;o--)
  m /= o;

How can I convert a combination function to permutation function?

2 Answers2