Help needed to solve combinatorics problem. I have been revisiting my old probability courses and I found a problem, which I can't figure out how to solve or at least what I get differs from the answer in the book. 
The problem reads as follows:
Given 8 balls i.e. 5 white 1 black, 1 blue and 1 red, find how many sequences of 6 objects can be formed (ordering matters!). 
Hint1:
Only permutations and permutations with repetition might be used. This is taken from a German textbook (I have translated it to my best), and the question comes at the end of the chapter on permutations. Notice that variations $\frac{n!}{(n-k)!}$ and combinations $\frac{n!}{(n-k)!k!}$ are not covered yet.  
Hint2: 
The answer according to the textbook is 178. 
Please note, that all the answers given below use the combination formula to get the answer and they all differ from 178, which is supposed to be the right answer.  
 A: $3$ cases should be discerned:


*

*$3$ white balls are drawn, so $1$ choice for the other $3$ balls drawn.

*$4$ white balls are drawn, so $3$ choices for the other $2$ balls drawn.

*$5$ white balls are drawn, so $3$ choices for the other ball drawn.


If the order (of colors) in the group counts then there are $$6!\left[\frac{1}{3!}+\frac{3}{4!}+\frac{3}{5!}\right]=228$$
possible groups.
A: I have the same answer (or am making the same mistake) as d_e (228). We have the following cases: 
(I) Pick 5 White balls and one coloured ball. Let's pick black. Then we can pick the black ball first, second, ... or sixth. So six permutations; three colours in total, so 18 possibilities in this case.
(II) Pick 4 White balls and two coloured balls. If we have 4 White balls and one coloured ball, then we have five permutations by (I). The second coloured ball can go first, second,.. or sixth, so a total of 5 x 6 = 30 permutations. Choosing two colours is the same as leaving one out, so we can do this in three different ways for a Grand total of 90 possibilities.
(III) Pick three White balls and three coloured balls. Applying similar reasoning to (II), start with 3 White and 1 coloured, for four permutations. The second coloured can be picked first, second, ... fifth, for 5 x 4 permutations. And the last coloured ball can go first, second, ... sixth, for a total of 120 permutations. 
(I)+(II)+(III)= 18+90+120 = 228. 
I would be very interested in knowing where I have gone wrong, if 178 is the correct answer. 

Hi Alexander, I don't have enough reputation to add a comment yet, so I am editing my previous answer. Can you also post the question in the German original, please? I speak German and may be able to see if there is a subtle difference in the translation. 
A: need all combos for each case times # permutations divided by # permutations for identical balls.  can choose one, two or three colored balls.
$6![{{3\choose 1}\over{5!}}+{{3\choose 2}\over{4!}}+{{3\choose 3}\over{3!}}]=6![{3\over{5!}}+{3\over{4!}}+{1\over{3!}}]$ which simplifies to $228$
A: Your book is wrong and the answer is 228, as has been shown already.
It's $$3\times\frac{6!}{5!}$$ for when just one non-white is chosen
Plus $$3\times\frac{6!}{4!}$$ for when two non- whites are chosen
Plus $$\frac{6!}{3!}$$ for when three non-whites are chosen
Which makes a total of 228
A: Use exponential generating functions. The white balls give:
$$
1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \frac{z^5}{5!}
$$
Each of the other colors gives:
$$
1 + \frac{z}{1!}
$$
In all, we want:
$$
6! [z^6] \left( 1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \frac{z^5}{5!} \right) \cdot ( 1 + z)^3 
$$
My tame symbolic algebra system (maxima) tells me this is 228.
Alternatively, it can be computed by hand (ellipses are terms that don't affect the result):
$\begin{align}
6! [z^6] \left( e^z - \frac{z^6}{6!} - \dots \right) \cdot (1 + z)^3
  &= 6! \left( [z^6] e^z (1 + 3 z + 3 z^2 + z^3) - [z^6] \frac{z^6}{6!} \cdot (1 + \dots) \right) \\
  &= 6! \left(\frac{1}{6!} 
                + 3 \cdot \frac{1}{5!}
                + 3 \cdot \frac{1}{4!}
                + \frac{1}{3!}
                - \frac{1}{6!}
        \right) \\
  &= 228
\end{align}$
