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Three subsets (Posted on 2014-08-15) Difficulty: 3 of 5
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.

No Solution Yet Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

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another try | Comment 8 of 13 |
(In reply to solution -- now for real? by Charlie)

This time counting the sets A with four members including member zero:

Set A can have two members in either of 4 ways:
In each of these 4 possibilities, B and C can be divided with B getting 1, 2 or 3 out of the 8 remaining.

So far we get 4 * (C(8,1)+C(8,2)+C(8,3))

Set A can have four members in any of 13 ways:
B can have 1 or 2 of the remaining 6.

So we add in 13 * (C(6,1)+C(6,2))

Set A can have six members in one way:
Set B can have only 1 of the remaining four:

So add C(4,1)

When set A has 8 elements, B and C can't be divided unevenly without an empty set.

4 * (C(8,1)+C(8,2)+C(8,3)) + 13 * (C(6,1)+C(6,2)) + C(4,1) = 645

  Posted by Charlie on 2014-08-16 00:07:16
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