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?

posader86 2022-07-23 Answered
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?
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Answers (1)

encoplemt5
Answered 2022-07-24 Author has 15 answers
There are C ( 10 , 4 ) = 10 ! 4 ! 6 ! different sets of 4 numbers chosen among x 1 , x 2 , x 10 . Each number x i belongs to C ( 9 , 3 ) = 9 ! 3 ! 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:
1 C ( 10 , 4 ) 1 i < j < k < l 10 x i + x j + x k + x l 4 = 1 4 C ( 10 , 4 ) i = 1 10 C ( 9 , 3 ) x i = C ( 9 , 3 ) 4 C ( 10 , 4 ) i = 1 10 x i = 1 10 i = 1 10 x 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.
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New questions

Linear multivariate recurrences with constant coefficients
In the theory of univariate linear recurrences with constant coefficients, there is a general method of solving initial value problems based on characteristic polynomials. I would like to ask, if any similar method is known for multivariate linear recurrences with constant coefficients. E.g., if there is a general method for solving recurrences like this:
f ( n + 1 , m + 1 ) = 2 f ( n + 1 , m ) + 3 f ( n , m ) f ( n 1 , m ) , f ( n , 0 ) = 1 , f ( 0 , m ) = m + 2.
Moreover, is their any method for solving recurrences in several variables, when the recurrence goes only by one of the variables? E.g., recurrences like this:
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