A) 25
B) 215
C) 210
D) 220
Correct Answer: B
Solution :
A relation on a set A is said to be symmetric iff \[(a,b)\in A\Rightarrow (b,a)\in A,\forall a,b\in A\] Here A={3,4,6,8,9} Number of order pairs of\[A\times A=5\times 5=25\] Divide 25 order pairs of \[A\times A\]in 3 parts as follows: Part-A: (3,3), (4,4), (6,6), (8,8), (9,9) Part-5 : (3,4), (3,6), (3,8), (3,9), (4,6), (4,8),(4,9),(6,8),(6,9),(8,9) Part-C:(4,3),(6,3),(8,3),(9,3),(6,4),(8,4), (9,4), (8,6), (9,6), (9,8) In part -A, both components of each order pair are same. In part - B, both components are different but not two such order pairs are present in which first component of one order pair is the second component of another order pair and vice-versa. In part-C, only reverse of the order pairs of part -B are present i.e., if (a, b) is present in part - B, then (b, a) will be present in . part -C For example (3, 4) is present in part - B and (4,3) present in part -C. Number of order pair in. A, B and C are 5,10 and 10 respectively. In any symmetric relation on set A, if any order pair of part -B is present then its reverse order pair of part -C will must be also present. Hence number of symmetric relation on set A is equal to the number of all relations on a set D, which contains all the order pairs of part -A and part- B. Now\[n(D)=n(A)+n(B)=5+10=15\] Hence number of all relations on set\[D={{(2)}^{15}}\] \[\Rightarrow \]Number of symmetric relations on set\[D={{(2)}^{15}}\]
You need to login to perform this action.
You will be redirected in
3 sec
