In concert there are 20 singers will perform. For each singer there is a set of other singers that he wishes to perform after them.. In concert there are 20 singers will perform. For each singer $x$, there is a (possibly empty) set of other singers that the singer $x$ wishes to perform later than all the singers in that set. Can it happen that there are exactly 2010 orders of the singers such that all their wishes are satisfied?

Attempt:
First of all, the question is quite ambiguous. 
In my understanding: We have to find enough number of possible orders such that it is safe to make all the singers happy, because we don't know exactly what they want..but we can calculate all possible wishes. There is no such order that can make all possible wishes satisfied, for example: if singer-1 wants to perform later than 19 others and singer-2 also wants to perform at the end, then it is not possible to find such an order. So the 2010 orders must consists all possible wishes, for example: 1005 orders may fulfill half of all the possible wishes and the other 1005 orders may fulfill the other half.
Let us try with much more simpler case first, for example with only 3 singers. Let the singers be $s_{1}, s_{2}, s_{3}$. For $s_{1}$, then the set of other singers may be: $\{\}, \{s_{2}\}, \{s_{3}\}, \{s_{2}, s_{3}\}$ there are $2^{2}$ sets. Independently, same also for $s_{2}$ and $s_{3}$.
$$ s_{1} \rightarrow \{\}, \{s_{2}\}, \{s_{3}\}, \{s_{2}, s_{3}\} $$
$$ s_{2} \rightarrow \{\}, \{s_{1}\}, \{s_{3}\}, \{s_{1}, s_{3}\} $$
$$ s_{3} \rightarrow \{\}, \{s_{2}\}, \{s_{1}\}, \{s_{2}, s_{1}\} $$
One order that makes them happy is that for each singer the set is $\{\}$.
Now combining them, we have these possibilities:
We can imagine the concert-order have 3 spots: $ (p_{1}, p_{2}, p_{3})$, each $p_{i}$ is the ith performer. There are exactly $3 \cdot 2 = 6$ possible orders in the concert. One example of order would be 
$$(s_{1}, s_{2}, s_{3})$$
to be concise, this will satisfy the wishes if they are either
$$s_{1}: \{\}, s_{2}: \{\}, s_{3}: \{\} $$
or 
$$s_{1}:\{\},  s_{2}: \{\}, s_{3}: \{1\}$$
or
$$s_{1}: \{\},  s_{2}:\{\},  s_{3}: \{2\}$$
or
$$s_{1}:\{\},  s_{2}: \{\}, s_{3}: \{1,2\}$$
and several others.
So far it is not easy to solve the problem. Not quite clear how to approach the problem.
 A: First, let $(a,b)$ be a “valid” point if there exists a set of preferences/ wishes such that with $a$ singers we can have exactly $b$ orderings of the singers which satisfy the preferences.  In this notation, we would like to find out if $(20, 2010)$ is a valid point.
Considering smaller numbers first, obviously $(1, 1), (2, 1), (2, 2!), (3, 1), (3, 2), (3, 3), (3, 6)$ etc are all valid points for different sets of preferences, and it is not hard to find such points. So lets see how we can build larger valid sets given small ones.  It may be noted that given two valid sets $X=(m, M), Y=(n, N)$, we can construct $(m+n, MN)$ easily by adding a preference, which says all the singers in $X$ Wish to follow all singers in $Y$.  So now we have a way to build larger valid sets.
Now, from $2010=6\times5\times67$, all we need are some $a, b, c$ such that $(a, 6), (b, 5), (c, 67)$ are valid sets and $a+b+c=20$.  The tough one to tackle is $67$, but with some brute forcing, we can find $(13, 67)$ is valid.  This can be done by fixing $11$ in a column, and saying of the remaining two, one has to follow the first fixed five, and the other  precedes last fixed 3. 
Then the rest follows fast as $(4,5), (3, 3!)$ are easily obtained as valid.
