Splitting parties into committees I feel like this should be an extremely simple problem, but I can't quite figure it out. How many ways are there to split $2n + 1$ places in a committee among $3$ nonempty parties, such that a coalition of any $2$ parties constitutes a majority? Any help would be very appreciated. Thanks in advance!

Let $a$ and $b$ be the number of places in parties one and two.
All that we need is $a,b\leq n$ and $a+b\geq n+1$
When $a$ is $0$ there are $0$ possibilites of $b$
When $a$ is $1$ there is $1$ possibility for $b$ (It must be $n$).
When $a$ is $2$ there are $2$ possibilities for $b$ (It can be $n$ or $n-1$)
$\dots$
When $a$ is $n$ there are $n$ possibilities for $b$ (It can be any number between $1$ and $n$ including $1$ and $n$).
Therefore there are $1+2+3+\dots+n=\frac{n(n+1)}{2}$ ways to split the places into parties.
