How many ways are there to choose 5 groups of 30 workers? I have 90 workers and I have to do 6 groups so that the total number of workers in the groups is 30. In each group there must be at least one worker. Each group will be assigned a different task.
The first thing I have done is to count the ways of choosing 30 workers from a group of 90:
$C_{90,30}=\frac{90!}{30!60!}=\frac{90!}{30!60!}$
Now I'm going to calculate the possible ways to make 5 groups with the 30 workers. To each worker I assigned the number 0 and I used the number 1 to delimit each group. So the groups that I have are:
$PR_{35}^{30,5}=\frac{35!}{5!30!}$
So the total number of elections are:
$\frac{90!}{30!60!}·\frac{35!}{5!30!}$
The problem with this way of calculating it includes the possibility that there are two consecutive ones. This implies that you would have a group with 0 people and by hypothesis it can not be.
How do I not include the permutations that have two consecutive ones?
EDIT:
 I have thought about calculating the number of permutations that has 2 consecutive ones, 3 followed ones, 4 followed ones and the 5 followed ones and applying the Inclusion-exclusion principle. To calculate the permutations with two ones followed I treat them as if they were a single element and I get:
$PR_{34}^{30,3,1}$. 
In the same way I calculate the other cases and in the end I obtain that the possible ways of doing 5 groups with 30 people and no empty group are:
$PR_{35}^{30,5}-PR_{34}^{30,3,1}+PR_{33}^{30,2,1}-PR_{32}^{30,1,1}+PR_{31}^{30,1}$
Therefore, the solution to the problem is:
$C_{90,30}·(PR_{35}^{30,5}-PR_{34}^{30,3,1}+PR_{33}^{30,2,1}-PR_{32}^{30,1,1}+PR_{31}^{30,1})$
Its rigth?
 A: First select a subset of $30$ workers, which can be done in $\binom{90}{30}$ ways.  Then break the 30 workers into 5 non-empty sets. This can be done in $S(30,5)$ ways, where $S$ denotes a Stirling number of the second kind.  Now assign each of the 5 sets one of the tasks.  This can be done in $5!$ ways.  So in all, there are
$$\binom{90}{30}\; S(30,5)\; 5! \approx 6.23 \times 10^{44}$$
possible arrangements.
EDIT
By request, here is an alternative Stirling-free derivation, using the Principle of Inclusion / Exclusion (PIE).  
There are $\binom{90}{30}$ ways to select a subset of 30 workers. If we ignore the requirement that each of the 5 task groups must be non-empty, there are $N = 5^{30}$ ways to assign the 30 workers to the 5 tasks.  Let's say such a set of assignments has "Property $i$" if there is no one assigned to task $i$, for $1 \le i \le 5$.  To apply PIE, we total the number of arrangements with $j$ of the properties; call this sum $S_j$, for $1 \le j \le 4$.  Then
$$\begin{align}
S_1 &= \binom{5}{1} 4^{30} \\
S_2 &= \binom{5}{2} 3^{30} \\
S_3 &= \binom{5}{3} 2^{30} \\
S_4 &= \binom{5}{4} 1^{30} \\
\end{align}$$
By PIE, the number of arrangements with none of the properties, which is the number of arrangements in which every task has at least one person assigned to it, is $N-S_1+S_2-S_3+S_4$.  So taking into account the number of ways to select the subset of 30 workers, the number of possible arrangements is 
$$\binom{90}{30} (N-S_1+S_2-S_3+S_4)$$
which yields the same answer as the previous method using Stirling numbers.
