Number of $2n$-letter words using double $n$-letter alphabet, without consecutive identical letters How many words with $2n$ letters can be created if I have an alphabet with $n$ letters and each of the letters have to occur exactly twice in the word, but no two consecutive letters are equal?
Thanks!

We have 2n positions to fill, lets fill two positions with one letter.
We have n pairs of double letters.
For the first pair of double letters we have (2n-2)(2n-3) possible positions. This is because first we put the letter on a position not on the sides, and the other letter on the remaining positions that are not next to our letter. 
[edit] Now we proceed for the next letter: We have 2n-2 positions left to fill with another letter, use the above procedure. 
Keep doing this, in total n times, to get all words. It works because it doesnt matter where we place one letter (and his double) for the other letters. 
[EDIT] I have forgotten one thing:The words of the form a _ _ _ a.
Adding them we get:
For n to 0 get the product of:
(2n-2)(2n-3)
And add to this :
n times 'For n-1 to 2 get the product of:
(2n-2)(2n-3) '
This gives for n=3 : 30 words.
For n=4 : 840
