I was given the following interesting problem, of which a non-computation proof was desired, and I got some kind of proof that does not involve direct computation, but isn’t the proof the book (Arthur Engel’s Problem Solving Strategies) recommends, but anyway (It will be fun to generalise too) here is my (perhaps shortenable further) problem+proof:
Given the set of n elements S = {1,2,3,…,n}, where 1≤k≤n, construct all possible subsets of S having k elements. Now, by the Well Ordering Principle, each of these sets has a minimum value element. Prove that the average of all such (n,k) minimum value elements is (n+1)/(k+1).
This problem does at first glance seem to be challenging, but we can attack indirectly searching for patterns. What I did was to not bother about the arithmetic mean as such in the beginning, but about the sum, that is, the arithmetic mean multiplied by the number of elements.
This paragraph is an addendum after I noticed my proof doesn’t quite work in the visual sense for n=1 or n=2, with respective values of k, and also for the general case k=1, so to kind of preserve the post (which anyway shall see better methods that are more complete and encompassing) I shall merely comment that the proofs for those are straightforward: for n=1, there is only one subset {1}, and its AM is obviously 1. We can plug that into the given relation and indeed see that it holds true. For n=2, the subsets are {1} and {2} for k=1, and {1,2} for k=2; the respective values are 3/2 and 1 respectively, and they too fit the given relation for n>2. And finally for general n and k=1, the sets are {i}, i ranging from 1 to n, and so the AM is (n+1)/2, which as we shall see also holds. Anyway now to give the general proof as I did it for n,k>2, k≤n:
Let f(n,k) be the required AM of minima of k-cardinality subsets of {1,2,3,…,n}. Let g(n,k) denote the sum of the minima, so that
We shall examine this g(n,k) now. What I did was investigate what happens when n is replaced by n+1 (or should we say the set S is increased by an element n+1). Clearly the new sum g(n+1,k) will contain all possible minimum-elements of the k-subsets from the n elements 1,2,3,….,n, in addition to all k-subsets that contain n+1 as an element. The former sum is g(n,k), and the latter is g(n,k-1) because we can see easily that n+1 will never be the minimum element and so taking all k-subsets of {1,2,….,n+1} containing n+1 results in the same thing as taking all (k-1)-subsets of {1,2,…..,n}. So then we have the identity
………………………………..(1)
Having done that we seek other relations. It occured to me that the element 1 might have a key role to play, as in, it will appear as a minimum element whenever something else isn’t. So I decided to partition the different cases of k-sets summing up to g(n,k) into those that contain 1 and those that do not. Let us examine the k-sets where 1 is fixed. Since if 1 is a fixed element, the remaining k-1 elements will be selected out of n-1 elements, and the contribution to the sum of minimums from these sets will be the number of sets multiplied by 1, and so the sum of minimum elements of the k-subsets not containing 1 is
.
This sum is the set of all minimum elements of sets like {2,3,5,7,…}, {5,6,9….}, etc. each of k elements. Now we employ a simple trick that frankly I was happy to discover. I am not sure how to put it into words, but see this: the minimum element of {2,3,4} is 2, and that of {1,2,3} is 1. So why not be able to see that all k-subsets that don’t contain 1 are bijectable with all k-subsets containing 1 of the parent set {1,2,3,….,n-1}? It is like shifting all the elements backward by 1. This backward shift results in our above computed partial sum of non-1 k-subsets of {1,2,3….,n} to be equal to the partial sum of 1-included k-subsets of {1,2,3,….,n-1}, increased by the number of such sets. So we finally get
which reduces to, on applying derived relation (1) and a standard combinations’ identity,
So we get,
The required AM answer is hence
I note that it will be similar to find the value of an AM if the minimum is replaced by maximum.

