Let S = 1, 2, 3, 4, 5, 6 . Then the probability that a randomly...

Given:$S = \left\{1, 2, 3, 4, 5, 6\right\}$

Total number of onto functions from $S$ tto $S$$= 6 !$

Now, for $g\left(3\right) = 2g$\left(1\right)$:</p><p>$\begin{cases}g\left(3\right) & g\left(1\right) \\ 2 & 1 \\ 4 & 2 \\ 6 & 3\end{cases}$</p><p>$∴ $\begin{cases}g\left(3\right) = & 2g\left(1\right)\end{cases}$$can be defined in$3$ways.</p><p>$g\left(2\right), g\left(4\right), g\left(5\right)\text{\text{ and }}g\left(6\right)$can be anything and can be defined in$4 !$ways.</p><p>$∴$Number of onto functions for which \left[g\left(3\right) = 2g\left(1\right)\right] = 3 . 4 !$

Now required probability$= \dfrac{3 . 4 !}{6 !} = \dfrac{3 \times 4 !}{30 \times 4 !} = \dfrac{1}{10}$