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) - $39!$

Then

  • 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?

up vote 1 down vote accepted

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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.