How do you find the number of distinct arrangements of the letters in INSISTS?

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How do you find the number of distinct arrangements of the letters in INSISTS?

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Answer 1 · 2017-09-26T21:06:53

$\frac{7!}{2! 3!} = 420$ $(7!)/(2!3!) = 420$

Explanation:

If all of the $7$ $7$ letters were distinct then we could arrange them in $7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040$ $7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040$ different ways.

We can simulate that by giving the repeated letters subscripts...

$I_{1} N S_{1} I_{2} S_{2} T S_{3}$ $"I"_1 "N" "S"_1 "I"_2 "S"_2 "T" "S"_3$

For each possible arrangement, there are $2! = 2 \cdot 1 = 2$ $2! = 2 * 1 = 2$ different possible arrangements of the $I$ $"I"$ 's and $3! = 3 \cdot 2 \cdot 1 = 6$ $3! = 3 * 2 * 1 = 6$ different possible arrangements of the $S$ $"S"$ 's.

Hence the number of distinct arrangements of:

$INSISTS$ $"INSISTS"$

is:

$\frac{7!}{2! 3!} = \frac{5040}{2 \cdot 6} = 420$ $(7!)/(2!3!) = 5040/(2*6) = 420$

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How do you find the number of distinct arrangements of the letters in INSISTS?

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