$0,1,2,3,\ldots, n$ be $(n+1)$ numbers. In how many ways they can be summed up so that sum $>n$. $0$, $1$, $2$, $3$, $\ldots n$ be $(n+1)$ numbers. In how many ways they can be summed up so that sum $>n$.  Disregard the order of summation and no repetitions.
For example:


*

*If $n=2$, we have $(2,1)$ and $(2,1,0$) so the required answer is $2$

*If $n=3$, we have $(3,1)$, $(3,2)$, $(3,1,0)$, $(3,2,0)$, $(3,2,1)$ and 
$(3,2,1,0)$.  So the answer is $6$
Thank you in advance!
 A: The best I can do is a generating function.
The generating function for partitions into distinct parts using only integers $0$ to $n$ - allowing $0$ to count as a part only once - is
$$(1+x^0)(1+x^1)(1+x^2)(1+x^3)\cdots (1+x^n)=\prod_{j=0}^{n}(1+x^j)\tag{1}\label{1}$$
because a partition into distinct parts can consist of either $0$ or $1$ $0$s, $0$ or $1$ $1$s, $0$ or $1$ $2$s etc. These possibilities combine to produce a polynomial with general form
$$\prod_{j=0}^{n}(1+x^j)=\sum_{k=0}^{\binom{n+1}{2}}a_kx^k$$
where $a_k$ receives a contribution of $1$ from the product for every possible sum 
$$k=0k_0+1k_1+2k_2+\cdots +nk_n$$ 
where $k_i\in\{0,1\}$.
In your case you wish to find
$$\sum_{k=n+1}^{\binom{n+1}{2}}a_k=\sum_{k=0}^{\binom{n+1}{2}}a_k-\sum_{k=0}^{n}a_k$$
The first summation on the right can be found by substituting $x=1$ in our product $\eqref{1}$ (which gives $2^{n+1}$) and the second summation can be found by taking partial sums of the product $\eqref{1}$ by multiplying it by $(1-x)^{-1}=1+x+x^2+x^3+\cdots$ and then obtaining the coefficient of $x^n$ of the result using the $[x^n]$ operator

$$\sum_{k=n+1}^{\binom{n+1}{2}}a_k=2^{n+1}-[x^n]\frac{1}{1-x}\prod_{j=0}^{n}(1+x^j)\tag{Answer}$$

