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)
Set A can have two members in any 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 5 ways:
{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 5 * (C(6,1)+C(6,2))
Set A can have six members in one way:
{0,2,3,4,5,7}
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)) + 5 * (C(6,1)+C(6,2)) + C(4,1) = 477
Edited on August 15, 2014, 5:07 pm
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Posted by Charlie
on 2014-08-15 17:07:08 |
solution -- now for real?
