Question

If $A \setminus B = \{a, b\}$, $B \setminus A = \{c, d\}$ and $A \cap B = \{e, f\}$, then the set $B$ is equal to:

• A. $\{a, b, c, d\}$
• B. $\{e, f, c, d\}$
• C. $\{a, b, e, f\}$
• D. $\{c, d, a, e\}$
• E. $\{d, e, a, b\}$

Hint:

A Venn diagram is extremely helpful for set theory. $B$ is the entire right circle, which consists of the "moon" shape ($B \setminus A$) and the middle "football" shape ($A \cap B$). Simply sum the elements in these two regions.

Answer 1

The Correct Option is B

Concept: The set $B$ can be expressed as the union of its intersection with $A$ and the elements that belong exclusively to $B$. In set notation: \[ B = (B \setminus A) \cup (A \cap B) \]

Step 1: Identify the elements from the given subsets.
From the problem description

• $B \setminus A = \{c, d\}$ (elements in $B$ but not in $A$)
• $A \cap B = \{e, f\}$ (elements common to both $A$ and $B$)

Step 2: Combine the sets to find $B$.
By combining the exclusive elements of $B$ with the intersection:
\[ B = \{c, d\} \cup \{e, f\} \] \[ B = \{c, d, e, f\} \]

Step 3: Match with options.
The order of elements in a set does not matter. The set $\{e, f, c, d\}$ contains the exact same elements.