You are requested to create 3 disjoint
sets such that:

1. Their union is a set of 10 digits (i.e. integers from 0 to 9 inclusive).

2. The average value of the members of set

**A** is

**3.5**.

3. The number of members in

**B** is less than the number of members in

**C**.

How many distinct solutions are there?

Rem: No empty sets.

(In reply to

computer solution by Charlie)

Huh? Are we doing the same problem?

There are three sets, A and B and C. B has smaller cardinality than C, not A. A can have 2 or 4 or 6 elements. The number of solutions is much, much bigger than 9. Don't have time to count now, but here's one:

A = {0,7} B = {1,2,3} C = {4,5,6,8,9}

*Edited on ***August 15, 2014, 3:14 pm**