Let S = {1, 2, 3, 4, 5, 6}. Then the probabil | vedclass

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(A) The total number of onto functions from a set $S$ of $6$ elements to itself is $6! = 720$.We are given the condition $g(3) = 2g(1)$. Since the cod

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$\frac{1}{10}$

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(A) The total number of onto functions from a set $S$ of $6$ elements to itself is $6! = 720$.We are given the condition $g(3) = 2g(1)$. Since the codomain is $S = \{1, 2, 3, 4, 5, 6\}$,the possible pairs $(g(1), g(3))$ are $(1, 2), (2, 4),$ and $(3, 6)$. There are $3$ such pairs.For each pair,the remaining $4$ elements of the domain must be mapped to the remaining $4$ elements of the codomain such that the function remains onto. This can be done in $4! = 24$ ways.Thus,the number of favorable onto functions is $3 	imes 4! = 3 	imes 24 = 72$.The required probability is $\frac{3 	imes 4!}{6!} = \frac{3 	imes 24}{720} = \frac{72}{720} = \frac{1}{10}$.

Let $S = \{1, 2, 3, 4, 5, 6\}$. Then the probability that a randomly chosen onto function $g: S \to S$ satisfies $g(3) = 2g(1)$ is:

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