A) \[mn\]
B) \[{{2}^{mn}}\]
C) \[{{m}^{n}}\]
D) \[{{n}^{m}}\]
Correct Answer: D
Solution :
[d] The image of any given element in A can be any one of the image of an element in B. \[\therefore \] The m elements in A can be assigned images \[n\times n...\times n\] (m times)\[={{n}^{m}}\] ways \[\therefore \] Total mapping from A to B \[={{n}^{m}}\]
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