Krdytkyu

Let's say I have a set ${1, 2, ...x}$ and pick every distinct combination of length 3 (including repeated numbers) and create a new set by adding values equal to the product of the elements in each combination. For example, if x = 4 the new set would be:

$${1*1*1, 2*1*1, 2*1*2, 2*2*2, 3*1*1, ... , 4*4*3, 4*4*4}$$

These combinations are unique (that is, no value in the new set is repeated).

My goal is to find the sum of these values in terms of $x$. I already know the size of the new set in terms of $x$ which is why I ask for the average value (unless finding the sum is easier without the average and size).

Hopefully my explanation of the set was clear but if needed I can elaborate more.

Thank you kindly!

Here is a little different approach. I will write $n$ instead of $x$ as the variable. It's silly, but I can't get used to $x$ being an integer variable.

We seek to evaluate $T=T_1+T_2+T_3,$ where
$$begin{align}
T_3&=sum_{i=1}^n{i^3}\
T_2&=sum_{i=1}^nsum_{j=1\jne i}^n{ij^2}\
T_1&=sum_{i=1}^{n-2}sum_{j=i+1}^{n-1}sum_{k=k+1}^n{ijk}
end{align}$$
We will use the following well-known formulas:
$$begin{align}
S_3&=sum_{i=1}^n{i^3}={n^2(n+1)^2over4}\
S_2&=sum_{i=1}^n{i^2}={n(n+1)(2n+1)over6}\
S_1&=sum_{i=1}^n{i}={n(n+1)over2}
end{align}$$
Note first that $$T_3=S_3tag{1}$$
Then
$T_2=(1^2+2^2+cdots+n^2)(1+2+cdots+n)-(1^3+2^3+cdots+n^3).$ That is,
$$T_2=S_2S_1-S_3tag{2}$$
To compute $T_1,$ note that in the expression $(1+2+3+cdots+n)^3,$ each term of $T_3$ occurs once, each term of $T_2$ occurs $3$ times, and each term of $T_1$ occurs $6$ times so that
$$begin{align}
S_1^3 &= T_3+3T_2+6T_1\
T_1&={S_1^3-T_3-3T_2over6}=\
&={S_1^3-S_3-3S_2S_1+3S_3over6}=\
&={S_1^3-3S_2S_1+2S_3over6}tag{3}
end{align}$$
using $(1)$ and $(2)$.
Now, using $(1),(2),text{ and }(3),$ we get $$T={S_1^3+3S_2S_1+2S_3over6},$$ and substituting the formulas for $S_1,S_2,S_3$ and slogging through the algebra gives $$T=boxed{{n^2(n+1)^2(n+2)(n+3)over48}={n+1choose2}{n+3choose4}}$$
Sanity check: $n=2$ gives ${3choose2}{5choose4}=3cdot5=15,$ in agreement with your calculation.

Using saulspatz's advice, I found that the sum of the terms in the new set of size $x$ is equivalent to:

$$(1 + cdots + x)^3 - (1 + cdots + x)(2 + 3(1+2) + cdots + x(1 + cdots + x))$$ $$=
left( sum_{i=0}^xi right) *left[ left( sum_{i=0}^xi right)^2 - sum_{i=1}^x i *left(sum_{j=0}^i j right) right]$$

While I offer my gratitude to saulspatz for his/her help I do wonder if this could be expressed more simply.

Edit: I included a picture of an example of $x = 4$ to show this equivalency, where $S_A$ is the sum of the terms in the new set (what we are trying to find).