64 different arrangements are possible.
There are 6!/(3!*2!) = 60 arrangements.
40,320
There are 4 letters in IOWA, all non repeating. Arrangements are 4! or 24.
5! 5 * 4 * 3 * 2 *1 = 120 arrangements you take the number of letters in the words and make it a factorial.
Total arrangements are determined by the equation f(n) = n!, where n is the number of letters in the word, and n! is the factorial function, which is n*(n-1) ... *1. This word has 11! total arrangements. Distinguishable arrangements are determined by the equation f(n) = n!/(c1!*c2! ... *cn!), where the denominator is the product of the factorials of the count of each unique letter in the word. There is one "m". There are four "i"s. There are four "s"s. There are two "p"s. So: 11!/(4!4!2!1!) = 39916800/1152 = 34650 distinguishable arrangements
act
three
7 factorial
120?
There are 7 factorial, or 5,040 permutations of the letters of ALGEBRA. However, only 2,520 of them are distinguishable because of the duplicate A's.
64 different arrangements are possible.
There are 3240 arrangements.
6! = 6x5x4x3x2x1 = 720 arrangements
There are 6! = 720 different arrangements.
181440
There are 6!/(3!*2!) = 60 arrangements.