(a) Consider the three Es as a single unit and the two Ds as another single unit. This reduces the problem to arranging the units: {EEE, DD, C, I, V}. There are 5 units to arrange, so the number of arrangements is given by:

\(5! = 120\)

(b) First, find the total number of arrangements of the letters in DECEIVED. The word has 8 letters with repetitions of E and D. The total number of arrangements is:

\(\frac{8!}{2!3!} = 3360\)

Next, find the number of arrangements where all three Es are together. Treat the three Es as a single unit, reducing the problem to arranging {EEE, D, D, C, I, V}. The number of arrangements is:

\(\frac{6!}{2!} = 360\)

Therefore, the number of arrangements where the three Es are not all together is:

\(3360 - 360 = 3000\)