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.