Problem Analysis #15

The problem asks for the number of relations $R$ on a set $A$ of size 6 such that $R$ is reflexive, symmetric, and $|R| = 10$.

1. Reflexive: $R$ must contain all pairs $(x, x)$ for $x \in A$. Since $|A| = 6$, there are exactly 6 such pairs.
2. Symmetric: For every $(x, y) \in R$, $(y, x)$ must also be in $R$.
3. Size: $|R| = 10$.

Since the 6 diagonal pairs $(x, x)$ are mandatory, let $R_{diag} = \{(x, x) \mid x \in A\}$. The remaining elements in $R$ must be off-diagonal pairs $(x, y)$ with $x \neq y$. Let $k$ be the number of off-diagonal elements. Then $|R| = 6 + k = 10$, so $k = 4$. Because $R$ is symmetric, the off-diagonal elements must come in pairs: if $(x, y) \in R$, then $(y, x) \in R$. Since $x \neq y$, $(x, y)$ and $(y, x)$ are distinct. Thus, the 4 off-diagonal elements must form 2 symmetric pairs: $\{(x_1, y_1), (y_1, x_1)\}$ and $\{(x_2, y_2), (y_2, x_2)\}$. This corresponds to choosing 2 distinct sets of unordered pairs $\{x, y\}$ from $A$ (where $x \neq y$).

The number of possible unordered pairs of distinct elements from $A$ is $\binom{6}{2}$. Let $N = \binom{6}{2} = 15$. We need to choose exactly 2 distinct pairs from these 15 to form the 4 off-diagonal elements. The number of ways to do this is $\binom{15}{2}$.

Calculation: $\binom{15}{2} = \frac{15 \times 14}{2} = 105$.

I will verify this with a Python script.

[PYTHON] import math

def solve(): n = 6 # Number of possible pairs {x, y} with x != y num_pairs = math.comb(n, 2)

# We need to choose 2 such pairs to get 2*2 = 4 off-diagonal elements
# Total elements = 6 (diagonal) + 4 (off-diagonal) = 10
num_ways = math.comb(num_pairs, 2)

print(f"Number of pairs (edges in K6): {num_pairs}")
print(f"Number of ways to choose 2 edges: {num_ways}")

solve() [/PYTHON] [THOUGHT] The code output confirms the calculation. Number of pairs is 15. Number of ways to choose 2 edges is 105. The logic holds:

• Reflexive elements are fixed (6 elements).
• Remaining 4 elements must be symmetric off-diagonal pairs.
• This means selecting 2 distinct unordered pairs from the $\binom{6}{2}$ available pairs.
• $\binom{15}{2} = 105$.

Final Answer seems to be 105. [/THOUGHT]

\boxed{105}

The set $A$ has 6 elements. A relation $R$ on $A$ is a subset of $A \times A$. Since $R$ is reflexive, it must contain all 6 diagonal elements $(x, x)$ for each $x \in A$. Since $|R| = 10$, there are $10 - 6 = 4$ off-diagonal elements in $R$. Since $R$ is symmetric, if $(x, y) \in R$ with $x \neq y$, then $(y, x) \in R$. Thus, the off-diagonal elements must occur in pairs. Let the off-diagonal elements be generated by a set of unordered pairs $\{x, y\}$ with $x \neq y$. Each such unordered pair contributes 2 elements to the relation: $(x, y)$ and $(y, x)$. To get exactly 4 off-diagonal elements, we must choose exactly 2 distinct unordered pairs of distinct elements from $A$.

The number of possible unordered pairs of distinct elements from a set of size 6 is $\binom{6}{2} = 15$. We need to choose 2 of these 15 pairs. The number of ways to do this is: $\binom{15}{2} = \frac{15 \times 14}{2} = 105$

\boxed{105}