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) = 2g(1) is :
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Official solution
g(3) = 2g(1) can be defined in 3 ways
number of onto functions in this condition = 3 \[\times\] 4!
Total number of onto functions = 6!
Required probability = \[{{3 \times 4!} \over {6!}} = {1 \over {10}}\]