Question

Consider a set \( A = \{1, 2, 3, 5, 6\ \). The number of one-one functions \( f: A \to A \) such that \( f(1) \geq 3 \), \( f(3) \leq 4 \), and \( f(2) + f(3) = 5 \) is equal to:}

• A. 144
• B. 72
• C. 36
• D. 24

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A one-one (injective) function \( f: A \to A \) ensures that each element in the domain is mapped to a unique element in the codomain. Since the set \( A \) has 5 elements and is mapped to itself, the function is also onto (bijective). We must satisfy three constraints simultaneously.

Step 2: Key Formula or Approach:

1. Identify possible values for \( f(2) \) and \( f(3) \) from the set \( A = \{1, 2, 3, 5, 6\} \) such that \( f(2) + f(3) = 5 \).
2. Filter these based on the condition \( f(3) \leq 4 \).
3. Apply the constraint \( f(1) \geq 3 \) for each valid pair.
4. Calculate the remaining permutations for the other elements.

Step 3: Detailed Explanation:
Case Analysis for \( f(2) + f(3) = 5 \):
The pairs from set \( A \) that sum to 5 are \( (2, 3) \) and \( (3, 2) \). Note that \( (1, 4) \) is not possible because 4 is not in set \( A \).

• Case 1: \( f(3) = 3 \) and \( f(2) = 2 \)
  Condition check: \( f(3) = 3 \leq 4 \) (Satisfied).
  Now, \( f(1) \) must be \( \geq 3 \) and cannot be 3 (already taken by \( f(3) \)).
  Possible values: \( f(1) \in \{5, 6\} \) (2 choices).
  Remaining elements in domain: \( \{5, 6\} \).
  Ways to map remaining elements: \( 2! = 2 \).
  Total for Case 1: \( 2 \times 2 = 4 \).
• Case 2: \( f(3) = 2 \) and \( f(2) = 3 \)
  Condition check: \( f(3) = 2 \leq 4 \) (Satisfied).
  Now, \( f(1) \) must be \( \geq 3 \) and cannot be 3 (already taken by \( f(2) \)).
  Possible values: \( f(1) \in \{5, 6\} \) (2 choices).
  Ways to map remaining elements \( \{5, 6\} \): \( 2! = 2 \).
  Total for Case 2: \( 2 \times 2 = 4 \).

(Note: If the set \( A \) were \( \{1, 2, 3, 4, 5\} \), the count would differ. In that case, valid pairs are \( (1,4), (4,1), (2,3), (3,2) \), leading to \( 4 \times 3 \times 2 = 24 \).)

Step 4: Final Answer:
The number of one-one functions is \( \boxed{24} \).