recursive combination without repetition

def getCombos(chars, k):
    print("Start of getCombos('" + chars + "', " + str(k) + ")")
    if k == 0:
        # BASE CASE
        print("For getCombos('" + chars + "', " + str(k) + ") base case returns ['']")
        return ['']  # If k asks for 0-combinations, return '' as the selection of zero letters from chars.
    elif chars == '':
        # BASE CASE - TODO - why does this *have* to come second?
        print("For getCombos('" + chars + "', " + str(k) + ') base case returns []')
        return []  # An empty set has no combinations for any k.

    # RECURSIVE CASE
    combinations = []
    head = chars[:1]
    tail = chars[1:]
    print("For getCombos('" + chars + "', " + str(k) + ") part 1, get combos of tail '" + tail + "'")
    tailCombos = getCombos(tail, k - 1)
    print("For getCombos('" + chars + "', " + str(k) + ') part 1, combos of tail are', tailCombos)
    print('  Adding head', head, 'to tail combos:')
    for combo in tailCombos:
        print('  New combination', head + combo)
        combinations.append(head + combo)

    print("For getCombos('" + chars + "', " + str(k) + ") part 2, include combos from getCombos('" + tail + "', " + str(k) + ')')
    combinations.extend(getCombos(tail, k))

    print("For getCombos('" + chars + "', " + str(k) + ') results are', combinations)
    return combinations

#print('3-Combinations of "ABCD":')
print('Results:', getCombos('ABC', 2))