1. ## Combinatorics?

It is a rule in Gaelic that no consonant or group of consonants can stand immediately between a strong and weak vowel;the strong vowels being a, o, u and the weak vowels being e and i. Find the number of Gaelic words of $\displaystyle (n + 3)$ letters each, which can be formed of n consonants and the vowels a,e,o where no letter is repeated in the same word.

Spoiler:
$\displaystyle \frac{2(n + 3)!}{(n + 2)}$

How to solve it?

2. Hello fardeen_gen
Originally Posted by fardeen_gen
It is a rule in Gaelic that no consonant or group of consonants can stand immediately between a strong and weak vowel;the strong vowels being a, o, u and the weak vowels being e and i. Find the number of Gaelic words of $\displaystyle (n + 3)$ letters each, which can be formed of n consonants and the vowels a,e,o where no letter is repeated in the same word.

Spoiler:
$\displaystyle \frac{2(n + 3)!}{(n + 2)}$

How to solve it?
The $\displaystyle n$ consonants can be arranged in a line in $\displaystyle n!$ ways.

There are then $\displaystyle (n+1)$ 'slots' into which arrangements of vowels can be placed. We must not place the 'e' alone in a slot, or there would then be one or more consonants between the 'e' and another vowel. So, we'll assume first that the 'e' is placed with another single vowel, and the remaining vowel in a separate slot.

There are $\displaystyle (n+1)$ ways of placing the 'e' into a slot, and then 4 ways of choosing and positioning an 'a' or an 'o' in the same slot alongside the 'e'. There are then $\displaystyle n$ ways of placing the third vowel into a different slot. Total: $\displaystyle 4n(n+1)$. But in half of these arrangements the third vowel will be placed in a slot where there is a group of consonants between the 'e' and this third vowel. So the permitted number of ways of positioning the group of 2 vowels and a single third vowel is $\displaystyle 2n(n+1)$.

Then there are $\displaystyle 3!=6$ ways of arranging the 3 vowels into a single group, and $\displaystyle (n+1)$ ways of choosing a 'slot' for this group. Total: $\displaystyle 6(n+1)$.

There are therefore $\displaystyle 2n(n+1) + 6(n+1) = 2(n+1)(n+3)$ ways of positioning the vowels into the $\displaystyle (n+1)$ 'slots' formed by the $\displaystyle n$ consonants.

So the total number of arrangements is $\displaystyle 2(n+1)(n+3)n! = \frac{2(n+3)!}{n+2}$.