Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

I think I can construct 3 * 4 + 1 distinct binary relations.

So \[\#P(A\times B)=2^{\#A\cdot \#B}\]

So the number of binary relations I cant get is given by:
\[ 2^{12} \]

Is it right?