Distributing a cake cut with four vertical slices among three people 
A round cake was cut with a knife $4$ times vertically in such a way
  that it is cut to maximum number of pieces.Find the number of ways of
  distributing these cakes among three people such that everyone gets
  at-least one piece.

My attempt: Maximum number of regions formed by $n$ straight lines in a plane is given by $1+\sum_{i=1}^n (i)$,so here the maximum cuts possible is $11$ now to me it seems that the rest is to count number of surjections between two sets of cardinality $11$ and $3$ respectively, which is  $3! \times $StirlingS$2[11, 3]=3^{11} -3 \times 2^{11} + 3$.
But the answer says it is to be $3^{11} -3(2^{11}+1)$My instructor says that this is correct and he explains it as follows: 

The total number of ways of distributing these cake pieces to $3$
  people = $3^{11}$
This includes $3\times2^{11}$ ways of distributing the cake pieces in
  which one person will not get any cake pieces, and $3$ ways of
  distributing $11$ cake pieces in which only one person will get all
  the cake pieces.
Therefore,the required number of ways = $3^{11} -3(2^{11}+1)$.

But I am quite sure that I haven't committed any mistake in recognizing the model but  I couldn't find a flaw in his reasoning either, also I couldn't convince myself why the counting surjections is not working here?!
 A: I think the OP's answer is correct. As the OP notes, we only need to find the number of surjections $f : \{ \mathrm{Slice}_1, \mathrm{Slice}_2, \ldots, \mathrm{Slice}_{11} \} \to \{ \mathrm{Child}_1, \mathrm{Child}_2, \mathrm{Child}_3 \}$. This is equal to
$$
3! \cdot S(11,3)
$$
where $S(\cdot, \cdot)$ refers to Stirling numbers of the second kind (wikipedia link). This turns out to be $3^{11}-3 \cdot 2^{11}+3$. 
Another way to do this is using the principle of inclusion-exclusion(link). 


*

*Let $N_0$ be the number of functions without any restriction. This is equal to $3^{11}$. 

*Let $N_1$ be the number of functions $f$ such that a particular fixed element (say $\mathrm{Child}_1$) of the codomain is not in the range of $f$. This is equal to $2^{11}$, since you pick one of the remaining 2 children for each slice.

*Let $N_2$ be the number of functions such that 2 fixed elements (say $\mathrm{Child}_1$ and $\mathrm{Child}_2$) are not in the range. This number is, of course, equal to $1$ (give all slices to the lucky third child). 

*Let $N_3$ be the number of functions such that all the elements of the codomain are not in the range. This is equal to $0$. (After all, we are not throwing away cake.)
So, applying inclusion-exclusion, the number of surjective functions turns out to be:
$$
\binom{3}{0} N_0 - \binom{3}{1} N_1 + \binom{3}{2} N_2 - \binom{3}{3} N_3 = N_0 - 3N_1 + 3N_2.
$$
This confirms OP's answer. 
Notice that we are adding $3N_2$ rather than subtracting it, because a function of "level 2" in the above set of definitions is automatically also present in level 1 and level 0. My guess is  official solution did not do the accounting carefully. 
A: You are correct: the answer should be $3^{11}-3\times 2^{11} +3 = 171006$ not $3^{11}-3\times 2^{11} -3 = 171000$.
There are various other ways of finding this apart from noting these are surjections: one is the inclusion-exclusion principle where the signs alternate, so ${3\choose 3}3^{11} - {3\choose 2}2^{11} + {3\choose 1} 1^{11} - {3\choose 0}0^{11}=171006$.  
Another would be to find how many ways exactly two people get some cake.  This is $3\times (2^{11}-2)= 6138$ (three ways of choosing which person is left out, and you have to exclude the possibilities that only one of the remaining pair gets something).  So you get $3^{11} - 3\times(2^{11} -2) - 3\times1^{11}=171006$.
