Counting

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​.

ok we have the integers 0,1,2...,9

to make the sum equal to 4 we have options

4 + 0 = 4

3 + 1 = 4

2 + 2 = 4 (not possible as we cant have duplicates)

and that is all

so new we can divide this into cases.

case # 1 (4 and 0 are the largest 2 elements):

if 0 is second largest element we have no element less than it so the only subset possible is {0, 4} so 1 option

case # 2 (3 and 1 as 2 max elements):

now we obviously have 0 which we can add such that it does not mess with the max elements as any element >= 2 it would mess up the "second-largest = 1" so we have two options one with zero and one without zero {0,1,3}, {1,3}

so adding up the cases we have 3 satisfying subsets