A) \[\frac{11!}{{{(2!)}^{3}}}\]
B) 59
C) 110
D) 56
Correct Answer: A
Solution :
There are 1M, 3E, 1D, II, IT, 2R, 2A, 2N R. .E . . . . . . . . . . . . rest of 11 letters can be arranged in \[\frac{11!}{{{(2\,\,\,!)}^{3}}}\]You need to login to perform this action.
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