Let S={0,1,…,9}. We are looking for subsets of S that have at least two elements, and the sum of the two largest elements in the subset is 4.
Let the subset be A⊆S. Let the two largest elements in A be a and b, with a>b. We require a+b=4.
Let's list the possible pairs (a,b) such that a>b and a+b=4:
If a=4, then b=0. The pair is (4,0).
In this case, 4 and 0 must be in the subset.
All other elements in the subset must be less than b=0. However, there are no elements in S less than 0.
So, the only subset for this case is {4,0}. This subset has two elements, and 4+0=4. So, this is a valid subset. (1 subset)
If a=3, then b=1. The pair is (3,1).
In this case, 3 and 1 must be in the subset.
All other elements in the subset must be less than b=1. The only element in S less than 1 is 0.
So, for this pair, the subset must contain {3,1}. We can optionally include 0.
The possible subsets are:
{3,1}
{3,1,0}
Both of these subsets have at least two elements, and the two largest elements are 3 and 1, with a sum of 4. (2 subsets)
If a=2.5, not possible as elements are integers.
Notice that a must be at least 2. The largest possible value for a is 4 (since a+b=4 and b must be at least 0).
Also, b must be at least 0.
And a>b.
Let's re-examine systematically:
The two largest elements, a and b (a>b), must sum to 4.
Possible pairs (a,b) from S satisfying a+b=4 and a>b:
Case 1: a=4,b=0
The subset must contain {4,0}.
All other elements in the subset must be smaller than b=0. There are no such elements in S.
So, the only subset is {4,0}. This subset has two elements, and the sum of its two largest elements (4 and 0) is 4.
Number of subsets for this case: 1.
Case 2: a=3,b=1
The subset must contain {3,1}.
All other elements in the subset must be smaller than b=1. The only element in S smaller than 1 is 0.
So, the subset must contain {3,1}, and we can either include or exclude 0.
The possible subsets are {3,1} and {3,1,0}.
Both of these subsets have at least two elements, and the sum of their two largest elements (3 and 1) is 4.
Number of subsets for this case: 2.
We cannot have a=2 because if a=2, then b=2, but we require a>b.
Therefore, the total number of subsets satisfying the given properties is the sum of the subsets from each case:
Total subsets = (Subsets from Case 1) + (Subsets from Case 2)
Total subsets = 1 + 2 = 3.
The subsets are:
{4,0}
{3,1}
{3,1,0}
The final answer is 3.
