ace test series

Question

If $A=\left \{ 1,2,3 \right \}$, then number of relations possible on $A$, which are neither reflexive nor symmetric is _____________ 

Comments

$\text{Is it 27}?$

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how 27?

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it is wrong , i did wrong

Answer 1

A={1,2,3}

A $\times$ A ={ (1,1)(2,2)(3,3)(1,2)(2,1)(1,3)(3,1)(2,3)(3,2) } 

Reflexive Relation :- A relation R on a set A is said to be Reflexive if  (xRx)∀x∈A  

$\underbrace{(1,1)(2,2)(3,3) }_{n}$$\underbrace{(1,2)(2,1)(1,3)(3,1)(2,3)(3,2) }_{n^{2}-n}$

Now For Reflexive relation, there is only one choice for diagonal elements (1,1)(2,2)(3,3)

and For remaining n²-n elements, there are 2 choices for each.Either it can include in relation or it can't include in relation.

Total number of reflexive relation = $1*2^{n^{2}-n} =2^{n^{2}-n}$

Symmetric Relation:- A relation 'R' on set A is said to be symmetric if (xRy)   then   (yRx) ∀x,y∈A

$\underbrace{(1,1)(2,2)(3,3) }_{n}$$\underbrace{(1,2)(2,1)(1,3)(3,1)(2,3)(3,2) }_{n^{2}-n}$

For n diagonal elements (1,1)(2,2)(3,3) there are 2 choices for each.Either it can include in relation or it can't include in relation.

For remaining n²-n elements according to definition of symmetric relation we can form pairs of(1,2)(2,1) and (1,3)(3,1)and (2,3)(3,2).For each pair, there are 2 choices.Either it can include in relation or it can't include in relation.

Hence,Total Number of Symmetric Relation= 2ⁿ*$2^{\frac{(n^{2}-n)}{2}}$=$2^{\frac{n(n+1)}{2}}$

Now For Reflexive and Symmetric relation there are only one choices for diagonal elements (1,1)(2,2)(3,3) and

for remaining.we can form pairs of(1,2)(2,1) and (1,3)(3,1)and (2,3)(3,2).For each pair, there are 2 choices.Either it can include in relation or it can't include in relation.

Hence,Total Number of Reflexive and Symmetric Relation = $2^{\frac{(n^{2}-n)}{2}}$=$2^{\frac{n(n-1)}{2}}$

n(R)=$2^{n^{2}-n}$

n(S)=$2^{\frac{n(n+1)}{2}}$

n(R∩S)=$2^{\frac{n(n-1)}{2}}$

Total number of relation which is Reflexive or Symmetric n(R∪S)= n(R) +n(S)-n(R∩S)

                                     = $2^{\frac{n(n+1)}{2}}+2^{n^{2}-n}-2^{\frac{n(n-1)}{2}}$

Total number of relation which is neither Reflexive nor Symmetric =Total relation-Total number of relation which is Reflexive or Symmetric

                                                                                               $=2^{n^{2}} - (2^{\frac{n(n+1)}{2}}+2^{n^{2}-n}-2^{\frac{n(n-1)}{2}})$

Given  n=3                                                         

                                                                                             =$512-(64+64-8)$

                                                                                            =$392$