The total number of functions,

$f:\{1,2,3,4\} \rightarrow\{1,2,3,4,5,6\}$ such that $f(1)+f(2)=f(3)$, is equal to :

<p>Given, $f(1) + f(2) = f(3)$</p> <p>It means $f(1),f(2)$ and $f(3)$ are dependent on each other. But there is no condition on $f(4)$, so $f(4)$ can be $f(4) = 1,2,3,4,5,6$.</p> <p>For $f(1),f(2)$ and we have to find how many functions possible which will satisfy the condition $f(1) + f(2) = f(3)$</p> <p>Case 1 :</p> <p>When $f(3) = 2$ then possible values of $f(1)$ and $f(2)$ which satisfy $f(1) + f(2) = f(3)$ is $f(1) = 1$ and $f(2) = 1$.</p> <p>And $f(4)$ can be = 1, 2, 3, 4, 5, 6</p> <p>$\therefore$ Total possible functions $=1\times6=6$</p> <p>Case 2 :</p> <p>When $f(3) = 3$ then possible values</p> <p>(1) $f(1) = 1$ and $f(2) = 2$</p> <p>(2) $f(1) = 2$ and $f(2) = 1$</p> <p>And $f(4)$ can be = 1, 2, 3, 4, 5, 6.</p> <p>$\therefore$ Total functions $= 2 \times 6 = 12$</p> <p>Case 3 :</p> <p>When $f(3) = 4$ then</p> <p>(1) $f(1) = 1$ and $f(2) = 3$</p> <p>(2) $f(1) = 2$ and $f(2) = 2$</p> <p>(3) $f(1) = 3$ and $f(2) = 1$</p> <p>And $f(4)$ can be = 1, 2, 3, 4, 5, 6</p> <p>$\therefore$ Total functions $= 3 \times 6 = 18$</p> <p>Case 4 :</p> <p>When $f(3) = 5$ then</p> <p>(1) $f(1) = 1$ and $f(4) = 4$</p> <p>(2) $f(1) = 2$ and $f(4) = 3$</p> <p>(3) $f(1) = 3$ and $f(4) = 2$</p> <p>(4) $f(1) = 4$ and $f(4) = 1$</p> <p>And $f(4)$ can be = 1, 2, 3, 4, 5 and 6</p> <p>$\therefore$ Total functions $= 4 \times 6 = 24$</p> <p>Case 5 :</p> <p>When $f(3) = 6$ then</p> <p>(1) $f(1) = 1$ and $f(2) = 5$</p> <p>(2) $f(1) = 2$ and $f(2) = 4$</p> <p>(3) $f(1) = 3$ and $f(2) = 3$</p> <p>(4) $f(1) = 4$ and $f(2) = 2$</p> <p>(5) $f(1) = 5$ and $f(2) = 1$</p> <p>And $f(4)$ can be = 1, 2, 3, 4, 5 and 6</p> <p>$\therefore$ Total possible functions $= 5 \times 6 = 30$</p> <p>$\therefore$ Total functions from those 5 cases we get</p> <p>$= 6 + 12 + 18 + 24 + 30$</p> <p>$= 90$</p>