Correct Answer - Option 3 : Both I and II
Let U = {1, 2}
All Possible subsets of U = {ϕ, {1}, {2}, {1, 2}}
A = (x, X), x ∈ X and X ⊆ U
x can be only {ϕ, 1, 2}
When x = 1
X = (1, {1})
X = {1, {1, 2}}
When x = 2
X = {2, {2}}
X = {2, {1, 2}}
Therefore, total elements in A, |A| = 2 + 2 = 4.
Option 1:
|A| = n × 2n - 1 = 2 × 22 - 1 = 4
Option 2:
\(\left| A \right| = \mathop \sum \limits_{k = 1}^n k\left( {\begin{array}{*{20}{c}} n\\ k \end{array}} \right) = 1 \times \left( {\begin{array}{*{20}{c}} 2\\ 1 \end{array}} \right) + 2 \times \left( {\begin{array}{*{20}{c}} 2\\ 2 \end{array}} \right)\)
\(\left| A \right| = 2 + 2 = 4\)
Both the options are correct.
Important Points:
x = ϕ and X = ϕ is not considered since ϕ ∈ ϕ is not true.
Although we cannot generalize just from one example but in general both the cases always hold true for given conditions