Let A = \ 1, 2, 3, 4, 5\ and B = \ 1, 2, 3, 4, 5, 6\ . Then the...

Given,

$f ( 1 ) + f\left(2\right) + 1 = f\left(4\right)$

$⇒ f\left(1\right) + f\left(2\right) \le 5$ \left\{\text{as}f\left(4\right) \le 6\right\}$</p><p>Now taking cases we get,</p><p>Case \left(1\right), f\left(1\right) = 1 ⇒ f\left(2\right) = \left\{1, 2, 3, 4\right\} \to 4\text{mappings}$

Case \left(2\right), f\left(1\right) = 2 ⇒ f\left(2\right) = \left\{1, 2, 3\right\} \to 3\text{mappings}$</p><p>Case \left(3\right), f\left(1\right) = 3 ⇒ f\left(2\right) = \left\{1, 2\right\} \to 2\text{mappings}$

Case \left(4\right), f\left(1\right) = 4 ⇒ f\left(2\right) = \left\{1\right\} \to 1\text{mappings}$</p><p>And$f\left(5\right) \text{ and } f\left(6\right) \to 6\text{mapping each}$</p><p>So, total number of function will be$= \left(4 + 3 + 2 + 1\right) \times 6 \times 6 = 360$