Let $f:\{1,2,3,4\}\to\{1,2,3\}$ find the probability such that picking a function formed with these sets it verifies this equality $f(1)f(2)f(3)=6.$
What I thought:
We have 3 cases for $f(1)=\{1,2,3\}$ so $f(1)$ can be equal to $1$, $2$ or $3$.
If $f(1)=1\implies f(2)=\{2,3\}$ then if $f(2) = 2\implies f(3)=3$ and so on.. I counted $6$ possible combinations then the probability is $P=\frac {no. favorable}{no.possible}$ then $no.favorable=6$ and $no.possible=3^4=81$. Then $P=\frac 2{29}$. But it is not correct and I don't see where I'm wrong