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  • #24707
    Aritra
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    #24710
    swastik pramanik
    Participant

    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.

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