It's hard to tell without code
![Smile Smile](https://python-forum.io/images/smilies/smile.png)
but if I have to guess then along the road you sorted cows dictionary (row #12 in original code) and got list of tuples and then used for-loop with enumerate (row #14 where name 'key' means indice and 'value' means tuple in list) and then tried to add to said tuple:
>>> d = {'Maggie': 3, 'Herman': 7, 'Betsy': 9, 'Oreo': 6, 'Moo Moo': 3, 'Milkshake': 2,
... 'Millie': 5, 'Lola': 2, 'Florence': 2, 'Henrietta': 9}
>>> sorted(d.items())
[('Betsy', 9), ('Florence', 2), ('Henrietta', 9), ('Herman', 7), ('Lola', 2), ('Maggie', 3), ('Milkshake', 2), ('Millie', 5), ('Moo Moo', 3), ('Oreo', 6)]
>>> for key, value in enumerate(sorted(d.items())):
... print(key, value)
...
0 ('Betsy', 9)
1 ('Florence', 2)
2 ('Henrietta', 9)
3 ('Herman', 7)
4 ('Lola', 2)
5 ('Maggie', 3)
6 ('Milkshake', 2)
7 ('Millie', 5)
8 ('Moo Moo', 3)
9 ('Oreo', 6)
Regarding combinations with dictionaries - it's can be achieved quite easily. Below there is step-by-step approach (which can be condensed if required):
>>> d = {'Maggie': 3, 'Herman': 7, 'Betsy': 9, 'Oreo': 6, 'Moo Moo': 3, 'Milkshake': 2,
... 'Millie': 5, 'Lola': 2, 'Florence': 2, 'Henrietta': 9}
>>> combs = ({j: d[j] for j in i} for i in combinations(d, 2)) # generator of dictionaries to yield all pair combinations
>>> qualifying = (row for row in combs if sum(row.values()) < 10) # generator to yield those pairs which weight is less than 10
>>> sorted(qualifying, key=lambda x: sum(x.values()), reverse=True) # consuming piped generators and getting sorted list of dictionaries
[{'Maggie': 3, 'Oreo': 6},
{'Herman': 7, 'Milkshake': 2},
{'Herman': 7, 'Lola': 2},
{'Herman': 7, 'Florence': 2},
{'Oreo': 6, 'Moo Moo': 3},
{'Maggie': 3, 'Millie': 5},
{'Oreo': 6, 'Milkshake': 2},
{'Oreo': 6, 'Lola': 2},
{'Oreo': 6, 'Florence': 2},
{'Moo Moo': 3, 'Millie': 5},
{'Milkshake': 2, 'Millie': 5},
{'Millie': 5, 'Lola': 2},
{'Millie': 5, 'Florence': 2},
{'Maggie': 3, 'Moo Moo': 3},
{'Maggie': 3, 'Milkshake': 2},
{'Maggie': 3, 'Lola': 2},
{'Maggie': 3, 'Florence': 2},
{'Moo Moo': 3, 'Milkshake': 2},
{'Moo Moo': 3, 'Lola': 2},
{'Moo Moo': 3, 'Florence': 2},
{'Milkshake': 2, 'Lola': 2},
{'Milkshake': 2, 'Florence': 2},
{'Lola': 2, 'Florence': 2}]
If you look closely you can observe, that there are missing cows
![Big Grin Big Grin](https://python-forum.io/images/smilies/biggrin.png)
in dictionary as Henriette and Betsy are too heavy to make into pair. Of course there are repetitions as well.