A) n!
B) \[{{n}^{n}}\]
C) n(n - 1)
D) \[{{2}^{n}}\]
E) None of these
Correct Answer: B
Solution :
Explanation Option (b) is correct. The first ball can be placed in any one of the n cells in n ways. The second ball can also be placed in any one of the n cells in n ways. \[\therefore \] The first and second balls can be placed in n cells in \[n\times n\text{ }i.e.,\text{ }{{n}^{2}}\]ways. Similarly each of the rest balls can be placed in n ways. Hence the required number of ways \[=n\times n\times ...\text{ }\times n\]times = \[{{n}^{n}}.\]
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