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OpenStudy (anonymous):
Find the number of permutations in the word “freezer”
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OpenStudy (raden):
freezer ----> there are 7 letters r = 2 times e = 3 times so, number permutations = 7!/(2!3!) = 7.6.5.4.3.2.1/(2.1.3.2.1) = ...
OpenStudy (anonymous):
Raden is right!
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