# Sum of all the possible $\binom{N}{n}$ elements of a $N$ elements set

Let $$N$$, $$n$$ be two natural numbers, with $$N > n$$. We define the set A as $$A = \{ x_1, x_2, x_3, ..., x_N \}, x_1,...,x_N \in \mathbb{R}$$ If we randomly pick $$n$$ elements from the set $$A$$, we have $$N$$ choose $$n$$ combinations. In other words, is it possible to pick $$n$$ elements from $$A$$ and observe that their sum, over all the possible combinations satisfies the following relation:

$$\sum_{j = 1}^\binom{N}{n} \left( \sum_{j=1}^{n} x_j \right) = \binom{N-1}{n-1} \sum_{j=1}^{N} x_j$$

I've checked this relation to be true in several cases, but the notation is somewhat not convincing... Can somebody please help me formalize better this concept?

• Your notation: two $j$'s on the left with totally different functions, and you want to sum over subsets not integers up to $\binom{N}{n}$. See my answer for a more reasonable notation.. Commented Mar 4, 2020 at 16:14

Your sum notation is off, let $$S(N,n)$$ be the set of all $$n$$-size subsets of $$\{1,2, \ldots,N\}$$, there are $$\binom{N}{n}$$ many of them.
$$\sum_{S \in S(N,n)} \sum_{i \in S} x_i\tag{1}$$
and we can count it like this: for every $$x_i$$ (fixed), we can see that it is part of $$\binom{N-1}{n-1}$$ many distinct subsets from $$S(N,n)$$ and so added in a sum that many times, so contributes $$x_i \binom{N-1}{n-1}$$ to the total sum in $$(1)$$. As this holds for all elements $$x_i$$, $$i \in \{1,\ldots,N\}$$, the sum under $$(1)$$ equals
$$\binom{N-1}{n-1} \sum_{i=1}^N x_i$$