Let A = \ 1, 2, 3, 4, 5, 6\ . The number of one-one functions f: A...

Given $A = \{1, 2, 3, 4, 5, 6\}$. Since $f: A \rightarrow A$ is a one-one function, $f(x)$ takes distinct values from $A$ for each $x \in A$.

We are given the condition $f(2) + f(3) = 5$. The possible pairs for $(f(2), f(3))$ from the set $A$ are:
$(1, 4), (2, 3), (3, 2), (4, 1)$

Notice that for all these pairs, the condition $f(3) \leq 4$ is automatically satisfied.

We also have the condition $f(1) \geq 3$. Since $f$ is one-one, $f(1)$ cannot take the values already assigned to $f(2)$ and $f(3)$. Let us find the number of choices for $f(1)$ in each case:

Case 1: $(f(2), f(3)) = (1, 4)$
$f(1) \geq 3$ and $f(1) \notin \{1, 4\} \Rightarrow f(1) \in \{3, 5, 6\}$. (3 choices)

Case 2: $(f(2), f(3)) = (2, 3)$
$f(1) \geq 3$ and $f(1) \notin \{2, 3\} \Rightarrow f(1) \in \{4, 5, 6\}$. (3 choices)

Case 3: $(f(2), f(3)) = (3, 2)$
$f(1) \geq 3$ and $f(1) \notin \{3, 2\} \Rightarrow f(1) \in \{4, 5, 6\}$. (3 choices)

Case 4: $(f(2), f(3)) = (4, 1)$
$f(1) \geq 3$ and $f(1) \notin \{4, 1\} \Rightarrow f(1) \in \{3, 5, 6\}$. (3 choices)

In each of the 4 cases, there are exactly 3 choices for $f(1)$.

After assigning values to $f(1), f(2),$ and $f(3)$, there are exactly 3 elements left in the codomain to be assigned to $f(4), f(5),$ and $f(6)$. Since $f$ is one-one, these can be assigned in $3! = 6$ ways.

Total number of such functions = (Number of pairs) $\times$ (Choices for $f(1)$) $\times$ (Arrangements for remaining elements)
Total = $4 \times 3 \times 6 = 72$

Answer: $72$