How many numbers of permutations in the word embarrass?

Answer 1

The word "embarrass" has 9 letters, with the following frequency of letters: e (1), m (1), b (1), a (1), r (2), s (3). To find the number of permutations, we use the formula for permutations of multiset:

[ \frac{n!}{n_1! \cdot n_2! \cdot n_3! \cdots} ]

where ( n ) is the total number of letters and ( n_1, n_2, ) etc., are the frequencies of each letter. Thus, the number of permutations is:

[ \frac{9!}{1! \cdot 1! \cdot 1! \cdot 1! \cdot 2! \cdot 3!} = \frac{362880}{1 \cdot 1 \cdot 1 \cdot 1 \cdot 2 \cdot 6} = 5040 ]

Therefore, there are 5,040 distinct permutations of the letters in "embarrass."