Probability Part 2

∵ g : S → S
where S={1,2,3,4,5,6}
Total number of onto function = 6 !
Given condition g(3)=2g(1)
If g(1)=1, then g(3)=2
g(1)=2, then g(3)=4
and g(1)=3, then g(3)=6
There are only 3 possible cases
Number onto functions = 3 × 4 !
Required probability =

3×4!

6!

=

1

10