Select Page
• Author
Posts
• #24707
Aritra
Participant

#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.

You must be logged in to reply to this topic.