I am having trouble with some combinatorial question. Its not my field and the question is difficult for me. Any help will be appreciate.
Let $m_1,..., m_{{M}}$ be numbers such that $m_i \in \{0, 1, ..., 2m\}$ and $\sum_{i=1}^Mm_i=2m$.
Which possibilities and how many possibilities for the sum of half of $m_i$, i.e. for the sum $m_1+...+ m_{\frac{M}{2}}$ to be odd?
Thank you.
I have started with possibility that one of the $m_i, i=1,..., \frac{M}{2}$, say $m_1=2k-1$, for $k=1,..., m$ and the rest $m_2=...=m_{M/2}=0$. So, I can have $m$ such possibilities.
Now I am considering possibilities that, say, $m_3=...=m_{M/2}=0$ and $m_1=1$, $m_2=2k$, for $k=1,...,m-1$. Such possibilities is $m-1$.... and so on up to $m_1=2m$, I guess.
With three non-zero terms, I am stuck.... Plus, I have assumption that $\sum_{i=1}^Mm_i=2m$. And in my question I should consider sum of the half of $m_i$. I am confused.