Let S = 1 , 2 , 3 , 4 , 5 , 6 . Then the probability that a randomly chosen onto function g from S to S satisfies g 3 = 2 g 1 is :

Step-by-step solution

Given: S = 1 , 2 , 3 , 4 , 5 , 6 Total number of onto functions from S tto S = 6 ! Now, for g 3 = 2 g 1 : g 3 g 1 2 1 4 2 6 3 ∴ g 3 = 2 g 1 can be defined in 3 ways. g 2 , g 4 , g 5   &   g 6 can be anything and can be defined in 4 ! ways. ∴ Number of onto functions for which g 3 = 2   g 1 = 3 . 4 ! Now required probability = 3 . 4 ! 6 ! = 3 × 4 ! 30 × 4 ! = 1 10