Question

Find the number of permutations in the word “freezer".

Answer 1

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 2

Answer with explanation:

Number of alphabets in word "freezer"=7

f-----1

r-----2

e----3

z-----1

Permutation of Word "freezer"

`=(7!)/(2!*3!)\\\\=(7*6*5*4*3*2*1)/(3*2*1*2*1)\\\\=(7*6*5*4)/(2)\\\\=7*6*5*2\\\\=420`