for K=2 we can get 1980 only if we assume that pairs
A=12 B=12
A=12 B=12
are two different pairs like
A=11 B=12
A=12 B=11

I think u are not right.

Yes, see this programm:

#include <iostream>
#include <math.h>

using namespace std;

int Sum (int n)
{
    int s=0;
    while (n)
    {
        s+=n%10;
        n/=10;
    }
    return s;
}

int main()
{
    int k=0;
    for (int i=10; i<100; i++)
    {
        for (int j=10; j<100; j++)
        {
            if (Sum(i)+Sum(j)==Sum(i+j))
            {
                k++; cout<<i<<" "<<j<<"\n";
            }
        }
    }
            cout<<k;
    return 0;
}

It returns 1980, but you can easily see that if i=j this pair wasn't counted twice