# How many 4-letter “words” have no two consecutive letters identical - clarification needed on answer

Question: How many 4-letter "words" (any combination of 4 letters) have no two consecutive letters identical?

Answer: There are 26 choices for the first letter. Once the first letter is chosen, there are 25 choices for the second letter, then 25 choices for the third letter, then 25 choices for the fourth letter. Hence, there are $26 \cdot 25^3 = 406,250$ such four letter words.

But how can this be? With that, I could make ABBZ for example. But that has a pair of consecutive identical letters. Also, I could make ABBB, which has two pairs of consecutive identical letters. What am I missing? Where does my thinking go wrong?

Since no two consecutive letters can be the same, both ABBZ and ABBB are ruled out. Suppose the first letter chosen is A. The next letter you choose must be one of the $25$ letters in the alphabet that is not A. If the next letter that is selected is B so that you have AB, as in your examples, the third letter must be one of the 25 letters that is not B, which precludes both ABBZ and ABBB.
There are $26$ choices for the first letter. Once that letter is chosen, the second letter may be selected in $25$ ways since it cannot be the letter in the first position. The third letter can be chosen in $25$ ways since it cannot be the letter in the second position. The fourth letter can be chosen in $25$ ways since it cannot be the letter in the third position, which yields $26 \cdot 25 \cdot 25 \cdot 25$ possible four letter words in which no two consecutive letters are identical.