Question

Let $X=\{a,b,c,d,e}$ and $A=\{a,b,c,d\}$. Let $P=\{B:B\subseteq X \text{ and } A\setminus B=\{d\}\}$. Then the number of elements in the set $P$ is

• A. 1
• B. 2
• C. 4
• D. 3
• E. 8

Hint:

Logic Tip: When dealing with set difference $A \setminus B = C$, every element in $C$ must be absent from $B$. Every element in $A$ that is NOT in $C$ must be present in $B$. Elements outside of $A$ act as "free variables" that double the number of possible sets for each free element.

Answer 1

The Correct Option is B

Concept:
To evaluate the number of possible subsets for $B$, we must use the definition of set difference:

• $A \setminus B$ consists of elements that are in $A$ but NOT in $B$.

Step 1: Identify the given sets and the conditions for set B.
Given the universal set and set $A$: $$X = \{a, b, c, d, e\}$$ $$A = \{a, b, c, d\}$$ The problem states that $B \subseteq X$ and: $$A \setminus B = \{d\}$$
Step 2: Determine the status of elements from set A.
Since $A \setminus B = \{d\}$, the element $d$ is in $A$ but not in $B$. Thus: $$d \notin B$$ The elements $a, b, c$ belong to $A$ but do not appear in the result $A \setminus B$. This means they must have been subtracted out, which is only possible if they are present in set $B$. $$\{a, b, c\} \subseteq B$$
Step 3: Determine the status of the remaining elements in X.
The element $e$ is in $X$ but is not in $A$.
Because it is not in $A$, whether $e$ is included in $B$ or not has no effect on the set difference $A \setminus B$.
Therefore, $e$ is an optional element for set $B$.
Step 4: Construct the possible subsets for B and evaluate the size of P.
Based on the previous steps, $B$ must contain $\{a, b, c\}$ and may optionally contain $e$.
This gives us two possible sets: $$B_1 = \{a, b, c\}$$ $$B_2 = \{a, b, c, e\}$$ The set $P$ consists of these valid subsets: $$P = \{\{a, b, c\}, \{a, b, c, e\}\}$$ Thus, the number of elements in set $P$ is: $$|P| = 2$$