Q.Let $A = \{1, 2, 3\}$, $B = \{3, 4\}$ and $C = \{4, 5, 6\}$. Find

Concept: Cartesian Product — the set of all ordered pairs where the first element comes from the first set and the second from the second set.

Step 1: Compute $B \cap C$ and $B \cup C$

$B \cap C = \{3, 4\} \cap \{4, 5, 6\} = \{4\}$

$B \cup C = \{3, 4\} \cup \{4, 5, 6\} = \{3, 4, 5, 6\}$

Step 2: Find $A \times (B \cap C)$

$A \times \{4\} = \{(1,4), (2,4), (3,4)\}$

Step 3: Find $A \times B$ and $A \times C$

$A \times B = \{(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)\}$

$A \times C = \{(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)\}$

Step 4: Compute intersections and unions

$(A \times B) \cap (A \times C) = \{(1,4), (2,4), (3,4)\}$ …

The Cartesian product $A \times B$ is the set of all ordered pairs $(a,b)$ where $a \in A$ and $b \in B$. The key insight here is that the product operation distributes over set intersection and union — just like multiplication distributes over addition in arithmetic. This isn't a coincidence: both are binary operations that "pair" elements from two sets, and the distributive laws hold because the condition "$a$ is in $A$ and $b$ is in both $B$ and $C$" is logically equivalent to "$a$ is in $A$ and $b$ is in $B$" and "$a$ is in $A$ and $b$ is in $C$". We'll verify this explicitly.

Let's compute each part step by step.

1. Find $B \cap C$ first.

   $B = \{3,4\}$, $C = \{4,5,6\}$. The only common element is $4$, so $B \cap C = \{4\}$.

2. Compute $A \times (B \cap C)$.

   $A = \{1,2,3\}$ and $B \cap C = \{4\}$. Form all ordered pairs with first element from $A$ and second element from $\{4\}$:

3. Compute $A \times B$ and $A \times C$ separately. $A \times B$: pair each element of $A$ with each element of $B = \{3,4\}$:

$A \times C$: pair each element of $A$ with each element of $C = \{4,5,6\}$:

4. Find $(A \times B) \cap (A \times C)$. The common ordered pairs between the two sets are those where the second coordinate is in both $B$ and $C$ — that is, $4$. So we take all pairs with second element $4$:

This matches exactly the result from part (i), confirming $A \times (B \cap C) = (A \times B) \cap (A \times C)$.

5. Find $B \cup C$.

   $B = \{3,4\}$, $C = \{4,5,6\}$, so $B \cup C = \{3,4,5,6\}$.

6. Compute $A \times (B \cup C)$. …