The total number of functions, f: \ 1, 2, 3, 4\ → \ 1, 2, 3, 4, 5,...

Given, $f\left(1\right) + f\left(2\right) = f\left(3\right)</p><p>Now possible values$ of f\left(3\right)$are$2, 3, 4, 5 and 6$(it cannot be$1$as minimum value$f$\left(1\right)\text{\text{ and } }$ f\left(2\right) is $1$, so $1 + 1 = 2$ )

For each $k$, if $f\left(3\right)$ = k$then there will be \left(k - 1\right) set$ of values forf\left(1\right)$and f\left(2\right)$

For example if $f\left(3\right) = 3$ then possible value of $f\left(1\right) \text{ and } f\left(2\right)$ will be \left(1, 2\right)\text{or}\left(2, 1\right) ⇒ 2$values </p><p>Similarly for$f\left(3\right) = 4$, possible values will be \left(1, 3\right), \left(3, 1\right), \left(2, 2\right) ⇒3$ values.

And so on.

So number of possible combinations of $f\left(1\right), f\left(2\right) and f\left(3\right)$ are $1 + 2 + 3 + 4 + 5 = 15$

And for each value of $f\left(3\right) there will be$6$possible value of$f\left(4\right)$, so total cases will be$15 \times 6 = 90