Question

Let $A = \{1, 2, 3, 4, 5, 6\}$. The number of one-one functions $f: A \to A$ such that $f(1) \ge 3, f(3) \le 4$ and $f(2) + f(3) = 5$, is _________.

Answer 1

Correct Answer: 72

Step 1: Understanding the Question:
We need to find the number of injective (one-one) mappings from set $A$ to itself under specific constraints on values of $f(1), f(2), f(3)$.
Step 2: Key Formula or Approach:
Analyze the constraint $f(2) + f(3) = 5$ first, as it links two values. Then apply conditions on $f(3)$ and $f(1)$.
Step 3: Detailed Explanation:
The possible pairs for $(f(2), f(3))$ from set $A$ such that their sum is 5 are:
Case 1: $(f(2), f(3)) = (1, 4)$.
Check $f(3) \le 4$: $4 \le 4$ (True).
$f(1) \ge 3$ and $f(1) \notin \{1, 4\}$. So $f(1) \in \{3, 5, 6\}$. (3 choices).
Remaining 3 elements for remaining 3 spots: $3! = 6$ ways.
Total for Case 1: $3 \times 6 = 18$.

Case 2: $(f(2), f(3)) = (4, 1)$.
Check $f(3) \le 4$: $1 \le 4$ (True).
$f(1) \ge 3$ and $f(1) \notin \{4, 1\}$. So $f(1) \in \{3, 5, 6\}$. (3 choices).
Remaining spots: $3! = 6$ ways.
Total for Case 2: $3 \times 6 = 18$.

Case 3: $(f(2), f(3)) = (2, 3)$.
Check $f(3) \le 4$: $3 \le 4$ (True).
$f(1) \ge 3$ and $f(1) \notin \{2, 3\}$. So $f(1) \in \{4, 5, 6\}$. (3 choices).
Remaining spots: $3! = 6$ ways.
Total for Case 3: $3 \times 6 = 18$.

Case 4: $(f(2), f(3)) = (3, 2)$.
Check $f(3) \le 4$: $2 \le 4$ (True).
$f(1) \ge 3$ and $f(1) \notin \{3, 2\}$. So $f(1) \in \{4, 5, 6\}$. (3 choices).
Remaining spots: $3! = 6$ ways.
Total for Case 4: $3 \times 6 = 18$.

Summing all cases: $18 + 18 + 18 + 18 = 72$.
Step 4: Final Answer:
The total number of such one-one functions is 72.