Given the universal set U = whole numbers from 1 to 10, set A = multiples of 3 within U, set B = factors of 6 within U, and set C = even numbers within U, answer the following: a) What is the cardinal

$$
|U| = 10
$$
The cardinality for U is $\boxed{10}$.

• b) Illustrate the given information in a Venn-diagram. To illustrate in a Venn diagram, we first find the intersections: $A \cap B = \{3, 6\}$ $A \cap C = \{6\}$ $B \cap C = \{2, 6\}$ $A \cap B \cap C = \{6\}$

  Elements in regions:

  • Only A: $\{9\}$
  • Only B: $\{1\}$
  • Only C: $\{4, 8, 10\}$
  • $A \cap B$ only (not C): $\{3\}$
  • $A \cap C$ only (not B): $\emptyset$
  • $B \cap C$ only (not A): $\{2\}$
  • $A \cap B \cap C$: $\{6\}$
  • Elements in U but outside $A \cup B \cup C$: $\{5, 7\}$

  A Venn diagram would show a rectangle representing U, with three overlapping circles for A, B, and C. The elements would be placed in their respective regions as listed above.

• c) Is $A-(A \cup C) = (A \cap B)-(B \cup C)$, solve it.

$A-(A \cup C)$.
First, find $A \cup C$: $A \cup C = \{3, 6, 9\} \cup \{2, 4, 6, 8, 10\} = \{2, 3, 4, 6, 8, 9, 10\}$ Now, find $A-(A \cup C)$: $A-(A \cup C) = \{3, 6, 9\} - \{2, 3, 4, 6, 8, 9, 10\} = \emptyset$

$(A \cap B)-(B \cup C)$.
First, find $A \cap B$: $A \cap B = \{3, 6\}$ Next, find $B \cup C$: $B \cup C = \{1, 2, 3, 6\} \cup \{2, 4, 6, 8, 10\} = \{1, 2, 3, 4, 6, 8, 10\}$ Now, find $(A \cap B)-(B \cup C)$: $(A \cap B)-(B \cup C) = \{3, 6\} - \{1, 2, 3, 4, 6, 8, 10\} = \emptyset$

Since LHS = $\emptyset$ and RHS = $\emptyset$, the equality holds.  
Therefore, $A-(A \cup C) = (A \cap B)-(B \cup C)$ is $\boxed{\text{True}}$.

• d) What are the groups $A \cup B \cup C$ and $B \cup C$? Write down with reason.