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

1 Answer
Sep 26, 2017

Answer:

#(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...

#"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 #"I"#'s and #3! = 3 * 2 * 1 = 6# different possible arrangements of the #"S"#'s.

Hence the number of distinct arrangements of:

#"INSISTS"#

is:

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