The letters of the word CONSTANTINOPLE are written on 14 cards, one of each card. The cards are shuffled and then arranged in a straight line. How many arrangements are there where no two vowels are next to each other?

1 Answer
George C.
Feb 5, 2016

#457228800#

Explanation:

CONSTANTINOPLE

First of all just consider the pattern of vowels and consonants.

We are given #5# vowels, which will split the sequence of #14# letters into #6# subsequences, the first before the first vowel, the second between the first and second vowels, etc.

The first and last of these #6# sequences of consonants may be empty, but the middle #4# must have at least one consonant in order to satisfy the condition that no two vowels are adjacent.

That leaves us with #5# consonants to divide among the #6# sequences. The possible clusterings are #{5}#, #{4,1}#, #{3,2}#, #{3,1,1}#, #{2,2,1}#, #{2,1,1,1}#, #{1,1,1,1,1}#. The number of different ways to allocate the parts of the cluster among the #6# subsequences for each of these clusterings is as follows:

#{5}: 6#

#{4,1}: 6xx5 = 30#

#{3,2}: 6xx5 = 30#

#{3, 1, 1}: (6xx5xx4)/2 = 60#

#{2, 2, 1}: (6xx5xx4)/2 = 60#

#{2, 1, 1, 1}: (6xx5xx4xx3)/(3!) = 60#

#{1,1,1,1,1}: 6#

That is a total of #252# ways to divide #5# consonants among #6# subsequences.

Next look at the subsequences of vowels and consonants in the arrangements:

The #5# vowels can be ordered in #(5!)/(2!) = 60# ways since there are #2# O's.

The #9# consonants can be ordered in #(9!)/(3!2!) = 30240# ways since there are #3# N's and #2# T's

So the total possible number of arrangements satisfying the conditions is #252*60*30240 = 457228800#

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