Counting symmetric relations

To count symmetric relations on a three-element set, separate diagonal and off-diagonal choices. The three diagonal pairs (1,1), (2,2), and (3,3) may each be independently included or omitted, giving 3 binary choices. For off-diagonal pairs, symmetry forces (i,j) and (j,i) to be chosen together. There are C(3,2) = 3 unordered pairs of distinct elements: {1,2}, {1,3}, and {2,3}. Each unordered pair again gives two choices, both ordered pairs included or both excluded. Thus the total number of independent binary decisions is 3 + 3 = 6, so the number of symmetric relations is 2^6. Option B, 2^9, treats all nine ordered pairs as independent and ignores symmetry; C uses the wrong base, and D omits the off-diagonal choices.

For a symmetric relation on an n-element set, each diagonal pair (a,a) can be chosen independently, giving n choices of whether to include it. For two distinct elements a and b, symmetry forces (a,b) and (b,a) to be included together or excluded together. The number of unordered pairs of distinct elements is n(n−1)/2, so the total number of independent binary choices is n + n(n−1)/2 = n(n+1)/2. With n=6, this exponent is 6×7/2 = 21. Therefore the number of symmetric relations is 2²¹, so option C is correct. Counting all 36 ordered pairs independently would violate symmetry, while the smaller exponents omit valid independent choices.