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1. What does this program print? ``

python
def choose(items, k, start, current, result):
    if len(current) == k:
        result.append(current.copy())
        return
    for i in range(start, len(items)):
        current.append(items[i])
        choose(items, k, i + 1, current, result)
        current.pop()


items = [1, 2, 3, 4]
result = []
choose(items, 3, 0, [], result)
print(result)
print(f"count = {len(result)}")

``

[[1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4]] count = 3[[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]] count = 6[[1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4]] count = 4[[3, 2, 1], [4, 2, 1], [4, 3, 1], [4, 3, 2]] count = 4

2. What does this program print? ``

python
def choose(items, k, start, current, result):
    if len(current) == k:
        result.append(current.copy())
        return
    for i in range(start, len(items)):
        if i > start and items[i] == items[i - 1]:
            continue
        current.append(items[i])
        choose(items, k, i + 1, current, result)
        current.pop()


items = [2, 1, 1, 3]
items.sort()
result = []
choose(items, 2, 0, [], result)
print(result)
print(f"count = {len(result)}")

``

[[1, 1], [1, 2], [1, 3], [2, 3]] count = 4[[1, 1], [1, 2], [1, 3], [1, 2], [1, 3], [2, 3]] count = 6[[1, 2], [1, 3], [2, 3]] count = 3[[1, 1], [1, 1], [1, 2], [1, 3], [2, 3]] count = 5

3. How many groups of 3 can you make from 6 items, when the order inside a group does not matter? Write the number.

Type your answer.

4. In the loop, after items[i] is chosen, the recursive call passes what new start index? Write it as an expression in i.

Type your answer.

5. In the duplicate-skipping rule, why is the check i > start used and not i > 0?

Because i > 0 is slower to compute on every pass of the loopBecause the input is never sorted before the loop runsBecause the first item a deeper loop tries must always be used, and with i > 0 the second 2 that makes [2, 2] would be skippedBecause start keeps the group from being longer than k

K-subsets