Use more than one for in the same list comprehension:
[(i, j, k)
for i in (11, 22, 33)
for j in (44, 55, 66)
for k in (77, 88, 99)]
This results in all possible permutations (a cartesian product), where enumeration starts on the right:
[(11, 44, 77), (11, 44, 88), (11, 44, 99),
(11, 55, 77), (11, 55, 88), (11, 55, 99),
(11, 66, 77), (11, 66, 88), (11, 66, 99),
(22, 44, 77), (22, 44, 88), (22, 44, 99),
(22, 55, 77), (22, 55, 88), (22, 55, 99),
(22, 66, 77), (22, 66, 88), (22, 66, 99),
(33, 44, 77), (33, 44, 88), (33, 44, 99),
(33, 55, 77), (33, 55, 88), (33, 55, 99),
(33, 66, 77), (33, 66, 88), (33, 66, 99)]
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