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DM - Grand Test -Q1


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+2 votes
What is the number of relations which are either symmetric or anti symmetric on the set A = {1,2,3} ?

(A) 272

(B) 264

(C) 274

(D) 262
asked Aug 30, 2020 by gbeditor-2 (72,840 points)
reshown Aug 31, 2020 by gbeditor-2

2 Answers

+1 vote
 
Best answer

We need to find  n(S) + n(A) - n(S \cap A).

The total number of symmetric relation possible on a set containing n elements is : n(S) = 2^{\frac{n(n+1)}{2}}

The total number of antisymmetric relations possible on a set containing n elements is : n(A) = 2^{n}*3^{\frac{n(n-1)}{2}}

The total number of relations which are both symmetric and anti symmetric is : n(A\cap S) = 2^{n} Since only possible elements is diagonal elements.

So we have , number of relations which are either symmetric or anti symmetric = 64 + 216 - 8 = 272

Hence Option A

answered Sep 8, 2020 by deepak-gatebook (226,240 points)
+1 vote
set A = {1,2,3}  here cardinality n=3

AxA ={(1,1) ( 2,2) ,(3,3) ,(1,2) ,(2,1) , (1,3), (3,1) ,(2,3) ,(3,2) }  total elements 9   n=9

so total possible relations 2^9 =512. (2^n)

now in symmetric relation if aRb is there then bRa must also be there so allowed pairs are (1,1) ( 2,2) ,(3,3)  {n diagonal pair)  and either both 1,2 or 2,1 are present or both are absent similalry for 1,3  3,1  and 2,3 and 3,2 .

none of them is must so empty relation will also be symmetric .

for n  diagonal pairs we have 2 choices for rest i.e (n^2-n) /2 pairs we again have 2 choices . so total number of symmetric relations are 2^n x2^(n^2-n)/2 since n=3  it will be 8x8=64 symmetric relations.

now in anti-symmetric relation if aRb and  bRa are there then  a=b  so allowed pairs are (1,1) ( 2,2) ,(3,3)  {n diagonal pair)  and for rest pairs like either 1,2  or 2,1 or none of two  should be there  so there are 3 choices . same with 1,3 3,1  and 2,3 3,2 pairs . so number of anti-symmetric relations are 2^n x3^(n^2-n)/2 =8x27=216

now since pairs like (1,1) ( 2,2) ,(3,3) are allowed in both symmetric and antisymmtric so number of common relations will be 2^3=8  

number of relations which are either symmetric or anti symmetric = 64+216-8 =272 option A
answered Sep 8, 2020 by (18,750 points)
Answer:

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