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https://math.answers.com/math-and-arithmetic/How_many_permutations_are_there_of_the_letter_factorisation

How many permutations are there of the letter factorisation? - Answers

The word "factorisation" consists of 13 letters, with the following frequency of letters: f (1), a (2), c (1), t (1), o (2), r (1), i (1), s (1), n (1). To find the number of distinct permutations, we use the formula for permutations of multiset: ( \frac{n!}{n_1! \times n_2! \times ... \times n_k!} ), where ( n ) is the total number of letters, and ( n_i ) is the frequency of each distinct letter. Therefore, the number of permutations is ( \frac{13!}{2! \times 2!} = \frac{6227020800}{4} = 1556755200 ).



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How many permutations are there of the letter factorisation? - Answers

https://math.answers.com/math-and-arithmetic/How_many_permutations_are_there_of_the_letter_factorisation

The word "factorisation" consists of 13 letters, with the following frequency of letters: f (1), a (2), c (1), t (1), o (2), r (1), i (1), s (1), n (1). To find the number of distinct permutations, we use the formula for permutations of multiset: ( \frac{n!}{n_1! \times n_2! \times ... \times n_k!} ), where ( n ) is the total number of letters, and ( n_i ) is the frequency of each distinct letter. Therefore, the number of permutations is ( \frac{13!}{2! \times 2!} = \frac{6227020800}{4} = 1556755200 ).



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https://math.answers.com/math-and-arithmetic/How_many_permutations_are_there_of_the_letter_factorisation

How many permutations are there of the letter factorisation? - Answers

The word "factorisation" consists of 13 letters, with the following frequency of letters: f (1), a (2), c (1), t (1), o (2), r (1), i (1), s (1), n (1). To find the number of distinct permutations, we use the formula for permutations of multiset: ( \frac{n!}{n_1! \times n_2! \times ... \times n_k!} ), where ( n ) is the total number of letters, and ( n_i ) is the frequency of each distinct letter. Therefore, the number of permutations is ( \frac{13!}{2! \times 2!} = \frac{6227020800}{4} = 1556755200 ).

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      The word "factorisation" consists of 13 letters, with the following frequency of letters: f (1), a (2), c (1), t (1), o (2), r (1), i (1), s (1), n (1). To find the number of distinct permutations, we use the formula for permutations of multiset: ( \frac{n!}{n_1! \times n_2! \times ... \times n_k!} ), where ( n ) is the total number of letters, and ( n_i ) is the frequency of each distinct letter. Therefore, the number of permutations is ( \frac{13!}{2! \times 2!} = \frac{6227020800}{4} = 1556755200 ).

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