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1) Number of arrangements of the word PERMUTATIONS,Total number of empty spaces are 12, so number of arrangements are 12! But `T` is repeated twice, so final arrangements are 12!/2!2) As there are even places and there are two `T` so their positions will be 6th and 7th, _ _ _ _ _ T T _ _ _ _ _So the first 5 letters can be arranged by 5! And last 5 letters by 5!Therefore, total arrangement are 5!×5!3) Let the letters `P` and `S` be fixed, so remaining places are 10, so total arrangements are 10! , but there are two `T` again, so final arrangements are 10!/2!
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