How many different permutations can you get from the letters in the word INVISIBILITY?

1 Answer
MathFact-orials.blogspot.com
May 21, 2017

#(12!)/(5!)=3,991,680#

Explanation:

If each of the 12 letters in Invisibility were different (let's say we had the letters A through L), we would say that any of the 12 letters could go first, then any of the remaining 11 letters could go second, and so on. That would then give us:

#12xx11xx10xx9xx8xx7xx6xx5xx4xx3xx2xx1= 12! = 479,001,600#

But there are multiples of the letter I. 5 of them, in fact. And so we need to divide out by #5!#, which is the number of ways the I can internally combine. That gives us:

#(12!)/(5!)=(12xx11xx10xx9xx8xx7xx6xxcancel(5!))/cancel(5!)=3,991,680#

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