c permutation and combination with frequency

I have 2 strings where I have saved fixed 20 characters and these are “A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T” and same 20 char in string 2. so I will get 400 combinations of 2 character sets like
AA,AB,AC,AD,AE,AF,……………AT
BA,BB,BC,BD,BE,BF,…………..BT
CA,CB,CC,CD,CE,CF……………CT

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



int main()
{
    int count = 0;
    char sequence1[30] = "ABCDEFGHIJKLMNOPQRST";
    char sequence2[30] = " ABCDEFGHIJKLMNOPQRST ";
    const char sequence3[30] = "AAAGACDF";    /*sequence taken from user which will be of length 31*/
    printf(" %s %s \n ",sequence1,sequence2);
    for (int i=0; i<strlen(sequence1); i++)
    {
        for(int j=0;j<strlen(sequence2);j++)
        {
              printf(" %c %c",sequence1[i],sequence2[j]);
            for(int k=0; k < strlen(sequence3)-1; k++)
            {
                if(sequence3[k] == sequence1[i] && sequence3[k+1] == sequence2[j])
                {
                    //cout<<"(equal)\t";
                    count++;
                }
            }
        }
    }

    printf("the number of matches are found %d ",count);
    getch();

}

This way we will get 400 combinations (Which program I have created successfully) but then user will put the value till 31 characters witch will be treated as 3rd string for E.g. “AAGACDF” now I have to check the frequency of user input in the sequence of 12,23,34,45,56,67,78,89 (2 CHAR SET) means AA,AA,AG,GA,AC,CD,DF and need to show the frequency
of user input

OUTPUT:

AA=2
AG=1
GA=1
AC=1
CD=1
DF=1

Edited by WingedPanther, 26 April 2010 - 03:37 PM.
add code tags (the # button)