Given:S=1,2,3,4,5,6
Total number of onto functions from S tto S=6!
Now, for g(3)=2g(1):
g(3)246g(1)123
∴g(3)=2g(1)can be defined in 3 ways.
g(2),g(4),g(5)\text{&}g(6) can be anything and can be defined in 4! ways.
∴Number of onto functions for which [g(3)=2g(1)]=3.4!
Now required probability=6!3.4!=30×4!3×4!=101