I don't know how I would approach this question:
All the letters of the word ENCUMBERANCE are arranged in a line. Find the total number of arrangements, which contain all the vowels in alphabetical order but separated by at least one consonant.
The vowels will be in the order: A E E E U
Now, first, we arrange the 7 consonants: 7!/2!2! (7 consonants being arranged, N and C are repeated twice)
Between each consonant, and before and after the consonants, we will have 8 slots we can place the vowels into (this is called the insertion method).
_ C _ C _ C _ C _ C _ C _ C _
(where C is a consonant)
In these 8 slots, we need to put in the five vowels, in the order A E E E U, and 3 blank spaces.
[We can do this through the stars and bars method, where we have 5 stars (A E E E U) and 3 bars, for example:
* - * * - - * *
which is the same thing as:
A _ E E _ _ E U
or
- * - * * * - *
which is the same as:
_ A _ E E E _ U
The number of arrangements for this is 8!/3!5!]
Alternatively, we are simply choosing 3 slots to be blank, so arrangements = 8C3.
Multiplying the number of arrangements for the consonants and the vowels gives us:
(7!/2!2!) * (8!/3!5!) = 70560
(I hope this is right)
