(simple problem) finding the distribution for given constraints For even number $N$, I have three constraints to be satisfied for $C_{n}$ as follows. 
\begin{align}
&1)~\sum_{n=0}^{N}C_{n}=1\\
&2)~\sum_{n=0}^{N}n~C_{n}=\frac{N}{2}\\
&3)~~C_{n}=C_{N-n} ~~\text{for}~n\le\frac{N}{2}
\end{align}
Is it possible to find the distribution of $C_{n}$ that satisfies the above equations? How can I find the distribution? 
 A: Long answer: Your question is equivalent to asking for solutions of the matrix equation $\left(\begin{array}{ccccccccc} 1 & 1 & \ldots & & & &\ldots & 1 &1 \\ 1 & 2 & 3 & \ldots & && n-2 & n-1 & n \\
1 & 0 & 0 & \ldots& &\ldots & 0 & 0 &-1 \\
0 & 1 & 0 &\ldots& &\ldots  & 0  &- 1 & 0 \\
 & & & \vdots &&\vdots & & & \\
0&\ldots & & 1 & 0& -1 & &\ldots & 0\end{array}\right)C=\left(\begin{array}{c}1 \\ \frac{N}{2} \\ 0 \\ \vdots \\ 0\end{array}\right)$. 
You have $\frac{N}{2}+2$ constraints for $N+1$ variables, so you will have a range of solutions. However if by distribution you mean that the $C_n$ are the pmf of some distribution then you also need all of your $C_n$ to be non-negative which may narrow it down. Ultimately your answer will be of the form "any convex combination of the distributions $\{C^{(i)}\}_{i=1}^k$, where $k=\frac{N}{2}-1$". Eyeballing the equations though $C_{\frac{N}{2}}=1$ is a solution, as is $C_{N-n}=C_n=\frac{1}{2}$ for any $0\leq n<\frac{N}{2}$. And you can show that the rank of the matrix is $\frac{N}{2}+1$, and that these solutions are linearly independent, so this is the basis of solution vectors. 
Short Answer: You're basically asking for all of the symmetric distributions on $\{0,1,\ldots,N\}$ (your second condition is actually redundant given the first and third). What are the building blocks of such distributions? $C_{N-n}=C_n=\frac{1}{2}$ for all $0\leq n<\frac{N}{2}$, and $C_{\frac{N}{2}}=1$. Can any solution be made from a convex combination of these distributions? Yes. 
