The key insight for solving this problem is to use dynamic programming to build up the solution from smaller subproblems.

Let's define dp[i] as the number of ways to form the sum i using the given numbers in nums.

For each number num in nums, if i >= num, then dp[i] += dp[i - num]. This is because we can form the sum i by adding num to any combination that forms the sum i - num.

The base case is dp[0] = 1, as there is one way to form the sum 0 (by not selecting any number).

To implement this, we can use a bottom-up approach:

1. Initialize an array dp of size target + 1 with all elements set to 0.
2. Set dp[0] = 1 as the base case.
3. For each i from 1 to target:
   • For each num in nums:
     • If i >= num, then dp[i] += dp[i - num].
4. Return dp[target], which represents the number of ways to form the sum target.

This approach correctly counts all possible combinations because the order of elements matters in this problem. For example, [1, 2] and [2, 1] are considered different combinations.

We can also solve this problem using a top-down approach with memoization, which is essentially a recursive solution with caching to avoid redundant calculations.

Let's define a recursive function combinationSum4(nums, target) that returns the number of ways to form the sum target using the numbers in nums.

The recurrence relation is:

combinationSum4(nums, target) = sum(combinationSum4(nums, target - num)) for all num in nums if target >= num

The base case is combinationSum4(nums, 0) = 1, as there is one way to form the sum 0 (by not selecting any number).

To avoid redundant calculations, we can use memoization to store the results of subproblems that we've already solved.

Here's how we can implement this approach:

1. Initialize a memo array or map to store the results of subproblems.
2. Define a recursive function combinationSum4(nums, target, memo) that returns the number of ways to form the sum target.
3. If target == 0, return 1.
4. If target < 0, return 0.
5. If target is already in memo, return memo[target].
6. Initialize count = 0.
7. For each num in nums:
   • Add combinationSum4(nums, target - num, memo) to count.
8. Store count in memo[target] and return count.

This approach also correctly counts all possible combinations, and it can be more intuitive for some people to understand compared to the bottom-up approach.

The previous approaches work well when all numbers in nums are positive, which is the case for this problem according to the constraints. However, if nums can contain negative numbers, we need to be careful about potential infinite loops in our recursive solution.

For example, if nums = [1, -1] and target = 1, we can have infinite combinations like [1, -1, 1, -1, 1, ...] that add up to target.

To handle this case, we can modify our top-down approach to include a maximum length for the combinations. This is not necessary for the given problem, but it's a good extension to consider for more general cases.

Here's how we can modify our top-down approach:

1. Initialize a memo array or map to store the results of subproblems.
2. Define a recursive function combinationSum4(nums, target, maxLength, memo) that returns the number of ways to form the sum target with at most maxLength numbers.
3. If target == 0, return 1.
4. If target < 0 or maxLength == 0, return 0.
5. If (target, maxLength) is already in memo, return memo[(target, maxLength)].
6. Initialize count = 0.
7. For each num in nums:
   • Add combinationSum4(nums, target - num, maxLength - 1, memo) to count.
8. Store count in memo[(target, maxLength)] and return count.

For the given problem, we can set maxLength to target (since we can't use more than target numbers if all numbers are at least 1), or we can use the bottom-up approach which naturally handles this case.