The question is, how many ways to rearrange letters of "to be or not to be that is the question" so, that we would get:

  • 1 8-letter word
  • 1 4-letter word
  • 2 3-letter words
  • 6 2-letter words

Words can be in any order, and of course doesnt have to mean anything. ie "ot eb ro question to be that is not the" is suitable variant.

I figure that first step would be to count all permutations of string(len 39) - $39!$


  • space count 9
  • t count 7
  • o count 5
  • b count 2
  • e count 4
  • r count 1
  • n count 2
  • h count 2
  • a count 1
  • i count 2
  • q count 1
  • u count 1
  • s count 2

And since letters can actually be in any order(lets forget spaces) - $39!/(7!5!2!4!1!2!2!1!2!1!1!2!)=39!/(7!5!2!4!2!2!2!2!)$

But spaces puzzle me, those can be in any order, but there is restriction of not putting space as a first letter, as a last letter and no doublespaces.

How to formalize those restrictions?


1 Answer 1


Count permutations of the letters separately, ignoring where the spaces are, and then multiply with the number of permutations of the word lengths multiset $\{2,2,2,3,2,2,4,2,3,8\}$.


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