$A =\{1,2,3\}$ and $B=\{a,b\}$

Based on the text, the number of relations between sets can be calculated using $2^{mn}$ where $m$ and $n$ represent the number of members in each set.

Given this, I calculated this number to be $2^{6}=64$ but this number seems too large.

Did I correctly calculate this value?


Yes, you did. There are $3 \cdot 2=6$ pairs of one element from $A$ and one from $B$. Each of these pairs can be in the relation or not, so you have six twofold choices that are independent. That gives $2^6=64$


The total number of relations that can be formed between two sets is the number of subsets of their Cartesian product.

For example: $$ n(A) = p\\ n(B) = q\\ \implies n(AXB) = pq\\ Number\ of\ relations\ between\ A\ and\ B = 2^{pq}\\ $$

Remember that if $n(T) = m$, then the number of subsets of set $T$ will be $2^{m}$

  • 1
    $\begingroup$ Welcome to the MSE community. Thanks for using MathJax, it will help the rest of read your contributions. Your answer might have been just a tiny bit better if you had started out answering the direct question with a direct, Yes. $\endgroup$ – Stephen Meskin Nov 21 '17 at 6:42

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