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

Sep 26, 2017

(7!)/(2!3!) = 420

#### Explanation:

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

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

${\text{I"_1 "N" "S"_1 "I"_2 "S"_2 "T" "S}}_{3}$

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

Hence the number of distinct arrangements of:

$\text{INSISTS}$

is:

(7!)/(2!3!) = 5040/(2*6) = 420