Number of One-One Functions satisfying given conditions

To find the total number of one-one functions $f: A \rightarrow A$, where $A = \{1, 2, 3, 4, 5, 6\}$, satisfying the given conditions, we analyze the constraints step-by-step.

Since $f$ is a one-one function on a finite set $A$ of 6 elements, $f$ is a bijection, meaning all image values $f(1), f(2), \dots, f(6)$ must be distinct elements of $A$.

Step 1: Determine possible values for $(f(2), f(3))$

We are given: $f(2) + f(3) = 5$ Since $f(2), f(3) \in \{1, 2, 3, 4, 5, 6\}$ and $f(2) \neq f(3)$, the possible pairs $(f(2), f(3))$ are:

1. $(1, 4)$
2. $(4, 1)$
3. $(2, 3)$
4. $(3, 2)$

We check the condition $f(3) \le 4$ for each pair:

• For $(1, 4)$, $f(3) = 4 \le 4$ (Valid)
• For $(4, 1)$, $f(3) = 1 \le 4$ (Valid)
• For $(2, 3)$, $f(3) = 3 \le 4$ (Valid)
• For $(3, 2)$, $f(3) = 2 \le 4$ (Valid)

All 4 pairs satisfy $f(3) \le 4$.

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Step 2: Determine possible values for $f(1)$

We are given that $f(1) \ge 3$, which implies: $f(1) \in \{3, 4, 5, 6\}$

Since $f$ is one-one, $f(1)$ cannot equal $f(2)$ or $f(3)$. Thus: $f(1) \in \{3, 4, 5, 6\} \setminus \{f(2), f(3)\}$

Let's find the number of choices for $f(1)$ in each of the 4 cases:

1. Case 1: $(f(2), f(3)) = (1, 4)$ $f(1) \in \{3, 4, 5, 6\} \setminus \{1, 4\} = \{3, 5, 6\} \implies 3 \text{ choices for } f(1)$

2. Case 2: $(f(2), f(3)) = (4, 1)$ $f(1) \in \{3, 4, 5, 6\} \setminus \{4, 1\} = \{3, 5, 6\} \implies 3 \text{ choices for } f(1)$

3. Case 3: $(f(2), f(3)) = (2, 3)$ $f(1) \in \{3, 4, 5, 6\} \setminus \{2, 3\} = \{4, 5, 6\} \implies 3 \text{ choices for } f(1)$

4. Case 4: $(f(2), f(3)) = (3, 2)$ $f(1) \in \{3, 4, 5, 6\} \setminus \{3, 2\} = \{4, 5, 6\} \implies 3 \text{ choices for } f(1)$

In every case, there are exactly $3$ choices for $f(1)$.

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Step 3: Assign values to the remaining domain elements

For any choice of $f(1)$, $f(2)$, and $f(3)$, exactly 3 distinct elements of $A$ have been assigned.

The remaining 3 elements of the domain, $\{4, 5, 6\}$, must be mapped to the remaining 3 unassigned elements of $A$ in a one-one manner. The number of ways to do this is: $3! = 6$

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Step 4: Calculate the total number of functions

For each of the 4 cases of $(f(2), f(3))$, the number of valid one-one functions is: $\text{Number of ways per case} = (\text{choices for } f(1)) \times 3! = 3 \times 6 = 18$

Summing over all 4 cases: $\text{Total number of functions} = 4 \times 18 = 72$