Question

The total number of symmetric relation that can be defined on the set 1, 2, 3, 4, 5, 6, 7 is
(A). ${2^{49}}$
(B). ${2^{7}}$
(C). ${7^{7}}$
(D). ${2^{28}}$

Answer 1

Hint: The total number of symmetric relations for a certain set totally depends upon the cardinal number of the set and is given by the formula ${2^{\dfrac{{n(n + 1)}}{2}}}$

Complete step by step answer:
We know that the total number of symmetric relation in a set is given by ${2^{\dfrac{{n(n + 1)}}{2}}}$ where n is the number of elements in the set.
So let us put this formula and in place of n we will put 7 as there are a total 7 elements in the given set.

\therefore {2^{\dfrac{{n(n + 1)}}{2}}}\\\ = {2^{\dfrac{{7(7 + 1)}}{2}}}\\\ = {2^{\dfrac{{7 \times 8}}{2}}}\\\ = {2^{7 \times 4}}\\\ = {2^{28}} \end{array}$$ So from here it is clear that option D is the correct option here. Note: A symmetric relation is a kind of binary relation where if (a,b) exists then (b,a) will also exist. It must be noted that many students make mistakes while putting the correct formula they often use the total number of reflexive relation in symmetric i.e., $${2^{n(n - 1)}}$$ in place of $${2^{\dfrac{{n(n + 1)}}{2}}}$$