TITLE: Four-letter word contains no two consecutive equal letters.
QUESTION [6 upvotes]: This is taken from the book on Combinatorics, by Daniel Marcus.
Request vetting:

A16: Find the probability that a four-letter word that uses letters from $A,B,C,D,E$ contains (a) no repeated letters; (b) no two consecutive equal letters.

(a)
Total (Inclusion) ways to form four-letter word from five letters: $5^4$.
Ways to have repetitions of $A$:
Two $A$: Choose two positions for $A$ in $\binom{4}2= 6$ ways. Rest two positions can be filled in : $4\times 3= 12$ ways. So, by product rule: $72$ words. Also, the same for other $4$ letters, leading to $72\times 5= 360$ words.
Three $A$: Choose three positions for $A$ in $\binom{4}3= 4$ ways. Rest one position can be filled in : $4$ ways. So, by product rule: $16$ words. Also, the same for other $4$ letters, leading to $16\times 5= 80$ words.
Four $A$: Choose all four positions for $A$ in $\binom{4}4= 1$ ways. Also, the same for other $4$ letters, leading to $5$ words.
Sum of above three cases:
$360+80+5= 445.$
Left are: $5^4 - 445= 180.$
$p= \frac{180}{625}$
(b) no two consecutive same letters' case.
Approach#1:
To get, need subtract from $5^4$  the chances of having two consecutive same letters. This is divided into cases:
Case 1: Position #1,2 are the same, and position #4 can match too :  $5\times 1\times 4\times 5= 100$ cases.
Case 2: position #2,3 are the same: $5\times 4\times 1 \times 4= 80$ cases.
Case 3: Position #3,4 are the same, and position #1 can match too:
$5\times 4\times 1\times 5=100$
So, get : $625- 280= 345$ cases.
Approach#2:
Alternatively, can 'directly' calculate by having $5$ choices for the first position, then $4$ choices, then again need eliminate one choice (of the second element) to have $4$ choices for the third position, and same for the fourth position. This leads to: $5\times 4^3= 320$ choices possible.
The second approach seems correct, but not clear why first approach is faulty.
Seems there are too many factors to consider in the first approach & hence error-prone (to miss some case(s) as here). So, second approach seems better.

REPLY [2 votes]: some of  your answers are incorrect, and some also use an unnecessarily complex approach.
(a) no. of words with no repeats is simply $P^5_4 = 5\cdot4\cdot3\cdot2 =120$
(b) no. of words with no two consecutive equal letters = $5\cdot4\cdot4\cdot4 = 320\quad$
You can then compute probabilities in each case by dividing by $5^4$