Notice that \(|A\times B|=mn\). Also, \(|C_i|=n\) for all \(i=1,2,3,\cdots ,m\). hence \(\sum_{i=1}^m |C_i|=\sum_{i=1}^m n=mn\). Similarly, we can prove that \(\sum_{j=1}^n |D_j|=mn\).

And hence, we get our desired result.