3 $[\quad$ Case Analysis $\quad]$

Given five different positive numbers that can be divided into two groups such that the sums of the numbers in these groups are equal. In how many ways can this be done?

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Let's take some partition that satisfies the condition of the problem and prove that there are no other partitions. If one of the groups contains only one number, then this number is equal to the sum of all the others, which means that in any other partition (specifically, in the group containing this number), the sum of the numbers will be greater.

Suppose there are two partitions, and in each of them the numbers are divided into groups of two and three numbers.

Consider the groups consisting of two numbers. If they do not intersect, then the remaining number is zero (since the sum of the numbers in each of these groups is half the sum of all the numbers). If some number is in both of these groups, then the remaining numbers must also be equal, that is, the partitions coincide.

## Answer

The only way.