**Sketch of the Proof:**

**int-avg subset** (just a name)

Note that one element subsets are by default int-avg subsets. They are n in number. Removing those elements from

Let X be the collection of all **int-avg** **subsets** S such that the average of S is contained in S

Y be the set of all **int-avg subsets** S such that the average of S is not contained in S.

Adding or deleting the average of a set to or from that set does not change the average.

This operation sets up a one-to-one correspondence between X and Y, so X and Y have the same cardinality. Since

**Comment**

What is the cardinality of

T_n – n is the number of non-singleton sets ……ar non- singleton sets always occur in pairs,er modhye there exists a S which contains the average.amr mne hoy this is a shorter solution.i was working it out…..