Problem Analysis #15

To solve the problem, we analyze the given condition:

The set $\mathcal{S}$ consists of all reflexive and symmetric relations $R$ on the set $A = \{a, b, c, d, e, f\}$ such that $|R| = 10$.

Step 1: Understand the Structure of Reflexive and Symmetric Relations

A relation $R$ on a finite set $A$ is a subset of $A \times A$. For this specific case:

• $|A| = 6$
• A reflexive relation must contain all pairs $(x, x)$ for each $x \in A$. There are exactly $|A| = 6$ such pairs.
• A symmetric relation satisfies: if $(x, y) \in R$, then $(y, x) \in R$.

Therefore, for $R$ to be both reflexive and symmetric:

• It must include all 6 diagonal elements $(x, x)$.
• The off-diagonal elements come in pairs: $(x, y)$ and $(y, x)$ for $x \neq y$.

Thus, the total number of elements in $R$ is:

where $m$ is the number of unordered pairs $\{x, y\}$ with $x \neq y$ such that both $(x, y)$ and $(y, x)$ are included in $R$.

Step flim: Solve for $m$ Given $|R| = 10$

We are told:

So, there are exactly 2 unordered pairs $\{x, y\}$ (with $x \neq y$) such that the corresponding symmetric pair $(x, y), (y, x)$ is included in $R$.

Step 3: Count the Number of Such Relations

The total number of unordered pairs $\{x, y\}$ with $x \neq y$ from a 6-element set is:

Each such pair corresponds to a unique symmetric pair of ordered pairs in $A \times A$. Therefore, choosing 2 such unordered pairs gives us the number of ways to construct the relation $R$ satisfying the conditions.

Hence, the number of such relations is:

--- with exactly 10 elements (6 from the diagonal, 4 from the symmetric off-diagonal pairs).

Final Answer

[THOUGHT] We need to count the number of reflexive and symmetric relations $R$ on a 6-element set with exactly 10 elements.

• A reflexive relation must contain all 6 diagonal pairs $(x,x)$.
• A symmetric relation includes both $(x,y)$ and $(y,x)$ for $x \neq y$.
• The number of unordered off-diagonal pairs is $\binom{6}{2} = 15$.
• Each such unordered pair contributes 2 ordered pairs to $R$.
• Total elements in $R$: $6 + 2k$, where $k$ is the number of chosen unordered off-diagonal pairs.
• Set $6 + 2k = 10 \Rightarrow k = 2$.
• The number of ways to choose 2 unordered pairs from 15 is $\binom{15}{2}$.

Compute $\binom{15}{2}$ using Python.

[PYTHON] import math math.comb( if there was a typo: math.comb(15,2) </PYTHON> [/PYTHON] 105

[THOUGHT] We need to count the number of reflexive and symmetric relations $R$ on a 6-element set with exactly 10 elements.

• A reflexive relation must contain all 6 diagonal pairs $(x,x)$.
• A symmetric relation requires that if $(x,y) \in R$ and $x \neq y$, then $(y,x) \in R$. So the non-diagonal elements come in pairs.
• The number of unordered pairs $\{x, y\}$ with $x \neq y$ is $\binom{6}{2} = 15$. Each such unordered pair corresponds to 2 ordered pairs in the relation.
• Let $k$ be the number of these 15 unordered pairs included in $R$. Then the total number of elements in $R$ is $6 + 2k$.
• We want $6 + 2k = 10 \implies 2k = 4 \implies k = 2$.
• The number of such relations is the number of ways to choose 2 unordered pairs from 15, which is $\binom{15}{2}$.

Now compute $\binom{15}{2}$ using Python.

[PYTHON] import math print(math.comb(15, 2)) [/PYTHON]

We consider a 6-element set. A reflexive and symmetric relation must contain all 6 diagonal pairs $(x, x)$, leaving $20$ non-diagonal ordered pairs, which form $15$ unordered pairs $\{(x, y), (y, x)\}$ for $x \neq y$. Let $k$ be the number of such unordered pairs included in the relation. Then the total number of elements in the relation is $6 + 2k$. Setting $6 + 2k = 10$ gives $k = 2$. The number of ways to choose 2 unordered pairs from the 15 possible is $\binom{15}{2} = 105$.

105

[THOUGHT] We need to count the number of reflexive and symmetric relations on a 6-element set with exactly 10 elements.

• A reflexive relation must contain all 6 diagonal pairs $(x, x)$.
• A symmetric relation requires that if $(x, y) \in R$ and $x \neq y$, then $(y, x) \in R$. So non-diagonal elements come in symmetric pairs.
• The number of unordered pairs $\{x, y\}$ with $x \neq y$ is $\binom{6}{2} = 15$. Each such pair corresponds to 2 ordered pairs in the relation.
• Let $k$ be the number of these 15 unordered pairs included in $R$. Then total elements in $R$ is $6 + 2k$.
• Set $6 + 2k = 10 \implies k = 2$.
• The number of such relations is the number of ways to choose 2 unordered pairs from 15, which is $\binom{15}{2} = 105$.

\boxed{105}