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 $\frac{3 	imes 4!}{6!} = \frac{3 	imes 24}{720} = \frac{72}{720} = \frac{1}{10}$.

$(A)$ $f: R \rightarrow R$ is such that $f(x)=px+q$ $(p \neq 0)$,$\forall x \in R$	$I.$ $f$ is neither one-one nor onto
$(B)$ $f: R \rightarrow R^{+} \cup\{0\}$ is such that $f(x)=x^2$,$\forall x \in R$	$II.$ $f$ is both one-one and onto
$(C)$ $f: N \rightarrow N$ is such that $f(n)=n^2+2n+3$,$\forall n \in N$	$III.$ $f$ is one-one but not onto
$(D)$ $f: R \rightarrow R$ is such that $f(x)=2(\cos ^2 5x+\sin ^2 5x)$ $\forall x \in R$	$IV.$ $f$ is onto but not one-one
	$V.$ $f$ is a constant function and also a bijection

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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