Let $$A = \{1, 2, 3, 4, 5, 6\}$$. 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 :

To solve for the number of one-one functions $$f: A \rightarrow A$$, we need to satisfy three specific conditions while ensuring no two elements in $$A = \{1, 2, 3, 4, 5, 6\}$$ map to the same image.

Since $$f: A \rightarrow A$$, the values for $$f(2)$$ and $$f(3)$$ must be distinct elements from the set $$\{1, 2, 3, 4, 5, 6\}$$. The pairs $$(f(2), f(3))$$ that sum to 5 are:

• Case 1: $$f(2) = 1, f(3) = 4$$
• Case 2: $$f(2) = 4, f(3) = 1$$
• Case 3: $$f(2) = 2, f(3) = 3$$
• Case 4: $$f(2) = 3, f(3) = 2$$

All four cases above satisfy $$f(3) \le 4$$. Now we calculate the number of choices for $$f(1)$$ for each case, keeping in mind that $$f$$ is a one-one function (so $$f(1)$$ cannot equal $$f(2)$$ or $$f(3)$$).

Total ways to assign $$f(1), f(2), \text{ and } f(3)$$:

$$3 + 3 + 3 + 3 = 12$$ ways.

For each of the 12 ways identified above, we have 3 elements left in the domain $$\{4, 5, 6\}$$ and 3 elements remaining in the codomain. Since the function is one-one, these 3 elements can be mapped in $$3!$$ ways.

$$3! = 3 \times 2 \times 1 = 6 \text{ ways}$$

Total number of one-one functions = (Ways to pick $$f(1), f(2), f(3)$$) $$\times$$ (Ways to arrange remaining elements)

$$\text{Total} = 12 \times 6 = 72$$

Final Answer: 72