Let , where is an odd prime. Find the number of -element subsets of the sum of whose elements is divisible by .
Attempt.
Let be the set of all the element subsets of . Let denote the sum of the elements of a member of . Define a relation on as: if , for . Its clear that is an equivalence relation. Choose such that . Let denote the equivalence class of under .
Claim: whenever
Proof. Let be such that (Such an exists).
If is odd put else put .
So in any case is odd and .
Now let .
Consider a set , where each element of is reduced mod if necessary.
One can easily show that 's are all distinct, thus .
Also , thus .
So from each element of we can produce one element of .
Also, since , it follows that two distinct elements of don't match to a same element of .
Replacing with in the above, the claim follows.
To arrive at the required answer I just need to show that . But here I am stuck.