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

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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Partial sums of geometric series

Full solution

Q. What is the total number of different

12

-letter arrangements that can be formed using the letters in the word INTERMITTENT?

\newline

Answer:

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 $I$ $0$ time, and $I$ $1$ appears $I$ $0$ 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 $9979200$ .

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