Respuesta :

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) = ...

Answer with explanation:

Number of alphabets in word "freezer"=7

f-----1

r-----2

e----3

z-----1

Permutation of Word "freezer"

    [tex]=\frac{7!}{2!*3!}\\\\=\frac{7*6*5*4*3*2*1}{3*2*1*2*1}\\\\=\frac{7*6*5*4}{2}\\\\=7*6*5*2\\\\=420[/tex]    

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