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 count the number of bijections (one-one functions from a finite set to itself) that satisfy three given constraints on the function values for inputs 1, 2, and 3.
Step 2: Key Formula or Approach:
The best strategy is to use a case-by-case analysis. We should start with the most restrictive condition, which is $f(2) + f(3) = 5$, as it links two values and has only a few possible solutions. For each case, we then apply the remaining conditions and use the multiplication principle.
Step 3: Detailed Explanation:
The domain and codomain are $A = \{1, 2, 3, 4, 5, 6\}$. The function $f$ must be one-one.
The condition $f(2) + f(3) = 5$ implies the pair $(f(2), f(3))$ can be one of the following (since $f(2) \neq f(3)$):
- (1, 4)
- (4, 1)
- (2, 3)
- (3, 2)
Let's analyze each case:
Case 1: $(f(2), f(3)) = (1, 4)$.
- Check condition $f(3) \le 4$: $4 \le 4$ is satisfied.
- Now, assign $f(1)$. Condition: $f(1) \ge 3$. Also, $f(1)$ cannot be 1 or 4 (already used). So, $f(1)$ can be chosen from $\{3, 5, 6\}$, giving 3 choices.
- The remaining 3 inputs (4, 5, 6) must map to the remaining 3 outputs. This can be done in $3! = 6$ ways.
- Total ways for this case: $3 \times 6 = 18$.
Case 2: $(f(2), f(3)) = (4, 1)$.
- Check condition $f(3) \le 4$: $1 \le 4$ is satisfied.
- Assign $f(1)$. Condition: $f(1) \ge 3$. Also, $f(1) \notin \{4, 1\}$. So, $f(1)$ can be chosen from $\{3, 5, 6\}$, giving 3 choices.
- Remaining assignments: $3! = 6$ ways.
- Total ways for this case: $3 \times 6 = 18$.
Case 3: $(f(2), f(3)) = (2, 3)$.
- Check condition $f(3) \le 4$: $3 \le 4$ is satisfied.
- Assign $f(1)$. Condition: $f(1) \ge 3$. Also, $f(1) \notin \{2, 3\}$. So, $f(1)$ can be chosen from $\{4, 5, 6\}$, giving 3 choices.
- Remaining assignments: $3! = 6$ ways.
- Total ways for this case: $3 \times 6 = 18$.
Case 4: $(f(2), f(3)) = (3, 2)$.
- Check condition $f(3) \le 4$: $2 \le 4$ is satisfied.
- Assign $f(1)$. Condition: $f(1) \ge 3$. Also, $f(1) \notin \{3, 2\}$. So, $f(1)$ can be chosen from $\{4, 5, 6\}$, giving 3 choices.
- Remaining assignments: $3! = 6$ ways.
- Total ways for this case: $3 \times 6 = 18$.
The total number of such functions is the sum of the ways in all valid cases:
Total = $18 + 18 + 18 + 18 = 72$.
Step 4: Final Answer:
The total number of such one-one functions is 72.