What is the total number of different 12-letter arrangements that can be formed using the letters in the word INTERMITTENT? Answer:

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Word Count Analysis: The word INTERMITTENT has 12 letters in total, with some letters repeating. We have the following frequency of each letter: I appears 2 times, N appears 2 times, T appears 3 times, E appears 2 times, R appears 1 time, and M appears 1 time.Permutations Formula: To find the total number of different arrangements, we use the formula for permutations of a multiset: $ \frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!} $, where $ n $ is the total number of items to arrange, and $ n_1, n_2, ..., n_k $ are the frequencies of each distinct item.Substitute Values: In this case, $ n = 12 $, $ n_1 = 2 $ for I, $ n_2 = 2 $ for N, $ n_3 = 3 $ for T, $ n_4 = 2 $ for E, $ n_5 = 1 $ for R, and $ n_6 = 1 $ for M. Plugging these into the formula gives us $ \frac{12!}{2! \cdot 2! \cdot 3! \cdot 2! \cdot 1! \cdot 1!} $.Calculate Factorials: Now we calculate the factorial of each number and the division: $ \frac{479001600}{2 \cdot 2 \cdot 6 \cdot 2} $.Simplify Division: Simplifying the division, we get $ \frac{479001600}{48} $, which equals 9979200.

What is the total number of different $12$ -letter arrangements that can be formed using the letters in the word INTERMITTENT? $\newline$ Answer:

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