My Python code that passed all test cases:

def solve(a):
    n1 = a.count(1) # no. of 1's in a
    n2 = a.count(2) # no. of 2's in a
    n = len(a)
    
    cnt_1_1 = n1*(n1 - 1)//2  # no. of (1,1) pairs
    cnt_2_2 = n2*(n2 - 1)//2  # no. of (2,2) pairs
    cnt_1_n = n1*(n - n1)    #  no. of (1,n) pairs
    
    return 2*cnt_1_1 + cnt_2_2 + cnt_1_n

Note that a,b ≥ 1, so a+b/ab quickly approaches 0. Because we take the floor of the function only a few value pairs will give non-0 results. Experimenting around, you'll see that 1,1 gives 1+1/1*1 = 2/1 = 2 and 1,n | n≥2, yields 1+n/n, which will yield 1 for all values n. Next, 2,2 yields 4/4 = 1, and 2,n yields 2+n/2n = 0 for n ≥ 3.

So, we need 2*nP2 | n = # of 1s, which is 2*n!/(n-2)!2! = 2n(n-1)/2 = n(n-1)

Also, for each 1, each non-1 will yield 1, so with m 1s and n other numbers, we'll need n*m (note than n = len(a)-m).

Lastly, we'll need nP2 | n = # of 2s, which, similar to above = n(n-1)/2

As for coding, I reccommend creating a counter (you could just do an array [m,n], then calculate the necessary values from that and total them.

Test case 11 is not correct rest everything works fine. Any suggestions

Very Simple . We have to care about the only 2 and 1. Because other numbers , like 3,4 will give (3+4)/(3*4)= 0 .So don't care about them .

All one(1) among them , make value 2 because (1+1)/(1*1) = 2.

All one(1) with other numbers, make value 1 because like for 3 (1+3)/(1*3)= 1

All two among them will make 1 . Because for 2 (2+2)/(2*2)=1 .

So only count them and give the result

My solution with comments line

https://github.com/joy-mollick/Problem-Solving-Solutions-Math-Greedy-/blob/master/HackerRank-Pairwise%20Sum%20and%20Divide.cpp

Simple question

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {int t;
            cin>>t;
            while(t--)
            {
                long int n;
                cin>>n;
                long long int a[n];
                long long int s=0;
                long int c1=0,c2=0;
                for(int i=0;i<n;i++)
                {
                    cin>>a[i];
                    if(a[i]==1)c1++;
                    if(a[i]==2)c2++;
                }
                s+=c1*(n-1);     
                s+=(c2*(c2-1))/2;
                cout<<s<<endl;
            }
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    return 0;
}