Question

Consider two sets $A=\{a,b,c\},B=\{e,f\}$. If maximum numbers of total relations from A to B; symmetric relation from A to A and from B to B are $l,m,n$ respectively, then the value of $2l+m-n$ is

• A. $212$
• B. $184$
• C. $240$
• D. $64$

Answer 1

The correct option is B $184$

We have $n\left(A\right)=3$ and $n\left(B\right)=2$.

Now cartesian product of $A$ and $B$ can contain maximum $3\times 2=6$ elements.

Since relation is the possible subsets of cartesian product.

So maximum number of relations from $A$ to $B$ be $\left(l\right)=2^6=64$.

Now maximum possible symmetric relation from $A$ to $A$ be $\left(m\right)=2^{\frac{3(3+1)}{2}}=2^6=64$ and that of from $B$ to $B$ be $\left(m\right)=2^{\frac{2(2+1)}{2}}=2^3=8$. [Using formula for number of symmetric relation]

Now $2l+m-n=128+64-8=184$.