If the number of elements in the sets G and A | vedclass

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Question

(A) Given that,$n(G) = 3$ and $n(A) = 4$.$(A)$ The number of functions from $G 	imes G$ to $G$ is $n(G)^{n(G 	imes G)} = 3^{(3 	imes 3)} = 3^9 = 19

Vedclass · Accepted Answer

$A-V, B-I, C-III, D-II$

Vedclass · Answer

(A) Given that,$n(G) = 3$ and $n(A) = 4$.$(A)$ The number of functions from $G 	imes G$ to $G$ is $n(G)^{n(G 	imes G)} = 3^{(3 	imes 3)} = 3^9 = 19683$. The number of bijective functions from $G 	imes G$ to $G$ is $0$ (since $n(G 	imes G) = 9 
eq n(G) = 3$). Thus,the number of non-bijective functions is $19683 - 0 = 19683$. So,$A ightarrow V$.$(B)$ The number of bijective functions from $A$ to $A$ is $n(A)! = 4! = 4 	imes 3 	imes 2 	imes 1 = 24$. So,$B ightarrow I$.$(C)$ The number of functions from $G$ to $G 	imes A$ is $n(G 	imes A)^{n(G)} = (3 	imes 4)^3 = 12^3 = 1728$. So,$C ightarrow III$.$(D)$ The number of surjective functions from $A$ to $A 	imes A$ is $0$ because $n(A) = 4$ and $n(A 	imes A) = 16$. Since $n(A) Therefore,the correct match is $A-V, B-I, C-III, D-II$.

List-$I$	List-$II$
$A$. The number of non-bijective functions from $G \times G$ to $G$	$I$. $24$
$B$. The number of bijective functions from $A$ to $A$	$II$. $0$
$C$. The number of functions from $G$ to $G \times A$	$III$. $1728$
$D$. The number of surjective functions from $A$ to $A \times A$	$IV$. $12$
	$V$. $19683$

List-$I$	List-$II$
$(A)$ $\sinh x$	$(I)$ Domain is $(-1, 1)$,even function
$(B)$ $\text{sech } x$	$(II)$ Domain is $[1, \infty)$,neither even nor odd function
$(C)$ $\tanh x$	$(III)$ Even function
$(D)$ $\text{cosech}^{-1} x$	$(IV)$ Range is $\mathbb{R}$,odd function
	$(V)$ Range is $(-1, 1)$,odd function

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