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?
Thanks in advance !
Best Answer
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: $$ \begin{align} {1\over C(10,4)}\sum_{1\le i<j<k<l\le10}{x_i+x_j+x_k+x_l\over4}& ={1\over 4C(10,4)}\sum_{i=1}^{10}C(9,3)x_i\\ &={C(9,3)\over 4C(10,4)}\sum_{i=1}^{10}x_i={1\over10}\sum_{i=1}^{10}x_i \end{align} $$ 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.