Number of arrangements with no consecutive letter the same I was learning the following questions in this site
question 1
question 2
which are indeed similar problems. But, one thing I observed is, in both of these problems, letters are repeating only two times. But, consider the following situation where one letter repeats two times and another three times.

How many arrangements of the letters in the word ABCDDEEE have no
  consecutive letter the same?

With similar logic, by applying inclusion-exclusion principle,
Answer must be
$$\dfrac{8!}{2! \times 3!}-\left(\dfrac{7!}{3!}+\dfrac{6!}{2!}\right)+5! =2280$$
But, I have checked with a program and answer appears to be $960$.
Could you please help me to understand why I am going wrong here? If anyone any general approach is there, please suggest.

Write $X$ for each of $A$, $B$, $C$ and leave a lot of space around these:
$\ \underline{\quad}X\underline{\quad}X\underline{\quad}X\underline{\quad}\  $.
Then consider the following cases:
(a) All three $E$s in the same slot. This requires the two $D$s  to separate the $E$s. Makes $4$.
(b) Two $E$s in one slot and one $E$ in another slot. This requires one $D$ to separate the $E$s. The remaining $D$ can be put in any of the now $6$ slots. Makes $4\cdot3\cdot 6=72$.
(c) One $E$ each in three of the four large slots. Then select two of the now $7$ slots for a $D$. Makes $4\cdot{7\choose2}=84$.
We now have $4+72+84=160$ cases in all. Replacing the $X$s by $A$, $B$, $C$ in any order brings us to $6\cdot160=960$ admissible strings.
