"To be or not to be" permutations

Question

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?

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 $\{2,2,2,3,2,2,4,2,3,8\}$.