Let X be a set of 5 elements. The number d of | vedclass

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(D) Let $n = 5$ be the number of elements in $X$. Each element $x \in X$ can be in one of the following four disjoint regions: $1. x \in A \cap B$ $2.

Vedclass · Accepted Answer

$201 \leq d$

Vedclass · Answer

(D) Let $n = 5$ be the number of elements in $X$. Each element $x \in X$ can be in one of the following four disjoint regions: $1. x \in A \cap B$ $2. x \in A \setminus B$ $3. x \in B \setminus A$ $4. x 
otin A \cup B$ Since $A 
eq \phi, B 
eq \phi$,and $A \cap B 
eq \phi$,we use the principle of inclusion-exclusion. Total number of pairs $(A, B)$ is $4^5 = 1024$. Let $S$ be the set of all pairs $(A, B)$. Let $P_1$ be the property $A = \phi$,$P_2$ be $B = \phi$,and $P_3$ be $A \cap B = \phi$. We want to find the number of pairs satisfying none of these properties. Using inclusion-exclusion,the number of valid pairs is $d = 4^5 - [|P_1| + |P_2| + |P_3|] + [|P_1 \cap P_2| + |P_1 \cap P_3| + |P_2 \cap P_3|] - |P_1 \cap P_2 \cap P_3|$. $|P_1| = 2^5 = 32$ (since $B$ can be any subset). $|P_2| = 2^5 = 32$ (since $A$ can be any subset). $|P_3| = 3^5 = 243$ (each element is in $A \setminus B, B \setminus A,$ or outside both). $|P_1 \cap P_2| = 1^5 = 1$ $(A=\phi, B=\phi)$. $|P_1 \cap P_3| = 2^5 = 32$ ($A=\phi, A \cap B = \phi \implies B$ can be any subset). $|P_2 \cap P_3| = 2^5 = 32$ ($B=\phi, A \cap B = \phi \implies A$ can be any subset). $|P_1 \cap P_2 \cap P_3| = 1^5 = 1$. $d = 1024 - (32 + 32 + 243) + (1 + 32 + 32) - 1 = 1024 - 307 + 65 - 1 = 781$. Wait,the provided solution logic in the prompt was incorrect. Re-evaluating: The number of pairs $(A, B)$ such that $A, B 
eq \phi$ and $A \cap B 
eq \phi$ is $781$. Since $781 > 201$,the correct option is $(D)$.

Let $X$ be a set of $5$ elements. The number $d$ of ordered pairs $(A, B)$ of subsets of $X$ such that $A \neq \phi, B \neq \phi, A \cap B \neq \phi$ satisfies:

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