In how many ways can $2t+1$ identical balls be placed in $3$ boxes so that any two boxes together will contain more balls than the third? 
In how many ways can $2t+1$ identical balls be placed in $3$ boxes so that any two boxes together will contain more balls than the third? 

I think we have to use multinomial theorem, but I cannot frame the expression!

Notation: A box $X$ has $x$ balls inside.
First of all, note that no box can contain less than $1$ and more than $t$ balls. Also, the sum of two boxes mustn't be lower than $t + 1$.
Let's say box $A$ has $1 \leq a \leq t$ balls. Box $B$ must have then at least $t - a + 1 \leq b \leq t$ balls. This gives you $a$ ways of filling box $B$. The remaining balls go to box $C$.
Therefore, since $a$ ranges from $1$ to $t$, you have $\sum_{a=1}^{t} a = \dfrac{t\cdot\left(t+1\right)}{2}$ ways of filling these boxes that satisfy the condition.
Edit: I am assuming that boxes are distinguishable from eachother, so that the distribution $\{t, t, 1\}$ differs from $\{t, 1, t\}$
