# Average of the Averages.

From $$10$$ numbers $$a,b,c,...j$$ all sets of $$4$$ numbers are chosen and their averages computed. Will the average of these averages be equal to the average of the $$10$$ numbers?

I tried analyzing smaller set of numbers but it became cumbersome and I couldn't reach to definite conclusion for this.

Can someone please help me with this? How can we derive at the answer for this? How can we prove this to be $$TRUE$$ or $$FALSE$$. Also if this is true then can it be generalized for $$N$$ numbers too like :-

From $$N$$ numbers $$a,b,c,...$$ all sets of $$n$$ numbers are chosen and their averages computed. Will the average of these averages be equal to the average of the $$N$$ numbers?

• When you take average from subset, then average of weighted averages gives same. May 15, 2021 at 12:04
• @zkutch : can you please elaborate that? I am sorry but I didn't get you. May 15, 2021 at 12:08
• I started writing, but Intelligenti pauca wrote it more quickly. May 15, 2021 at 12:20

There are $$C(10,4)={10!\over4!6!}$$ different sets of 4 numbers chosen among $$x_1,x_2,\dots x_{10}$$. Each number $$x_i$$ belongs to $$C(9,3)={9!\over3!6!}$$ such sets, because the other three numbers in the same set can be chosen in $$C(9,3)$$ different ways. Hence the average on all sets is: \begin{align} {1\over C(10,4)}\sum_{1\le i and both averages are the same.
This also works in general for the case of all sets of $$n$$ numbers chosen among $$N$$. The key is all numbers $$x_i$$ appear the same number of times in the final sum.
• Can you please explain "Each number belongs to $C(9,3)$ such sets." ? And can you please also answer my question in the last bit of the problem statement? May 15, 2021 at 12:21