Let X = {1, 2, 3, 4, 5}. The number of different ordered pairs (Y, Z) that can be formed such that \(Y \subseteq X\), \(Z \subseteq X\) and \(Y \cap Z\) is empty, is
\(S = \{1, 2, 3\}\), \(f : S \to S\) satisfies the property: \(\forall x \in S,\; f(f(x)) = f(x)\). How many different functions are there for \(f(x)\)?[Note: Can you generalize the result for \(S = \{1, 2, 3, \ldots, n\}\)?]
Let \(R = \{(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)\}\) be a relation on the set \(A = \{3, 6, 9, 12\}\). The relation is
Let X = \{1, 2, 3, \ldots, 12\} and N be the number of pairs \{A, B\} such that A ⊆ X, B ⊆ X, A ≠ B and A ∩ B = \{2, 3, 5, 7, 8\}. Then the value of N is
Consider the graph of \(y = f(x)\) with key points \((-5,-1)\), \((-3,2)\), \((-1,1)\), \((0,3)\), \((2,3)\), \((4,2)\) (approaching \(y=2\)), \((5,-1)\). Find the number of solution(s) of \(x\) satisfying \(f(f(x)) = 2\).