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 Three subsets (Posted on 2014-08-15)
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.

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 final (?) revision | Comment 11 of 13 |
(In reply to another try by Charlie)

Set A can have two members in either of 4 ways:
{0,7}{1,6}{2,5}{3,4}
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:
{0,1,4,9}{0,1,5,8}{0,1,6,7}{0,2,3,9}{0,2,4,8}{0,2,5,7}{0,3,4,7}{0,3,5,6}
{1,2,3,8}{1,2,4,7}{1,2,5,6}{1,3,4,6}{2,3,4,5}
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 9 ways:
{0,1,2,3,6,9}{0,1,2,3,7,8}{0,1,2,4,5,9}{0,1,2,4,6,8}{0,1,2,5,6,7}{0,1,3,4,5,8}{0,1,3,4,6,7}{0,2,3,4,5,7}{1,2,3,4,5,6}
Set B can have only 1 of the remaining four:

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)) + 9 * C(4,1) = 677

Edited on August 16, 2014, 10:41 am
 Posted by Charlie on 2014-08-16 10:41:22

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