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![Robson Possidonio](https://secure.gravatar.com/avatar/1c237572f79289d1d79fdd1982188d6f?s=96&d=https%3A%2F%2Fecs-static.teamtreehouse.com%2Fassets%2Fcontent%2Fdefault_avatar-445fbbabfc8dc9188fb5967fe43322ee0c3e0dd1e10f378bf8343784af5a13eb.webp&r=pg)
Robson Possidonio
977 Pointsone way to answer?
I think I've figured out another way to solve this task, but I keep getting an error. are alternative answers a no no?
musical_groups = [
["Ad Rock", "MCA", "Mike D."],
["John Lennon", "Paul McCartney", "Ringo Starr", "George Harrison"],
["Salt", "Peppa", "Spinderella"],
["Rivers Cuomo", "Patrick Wilson", "Brian Bell", "Scott Shriner"],
["Chuck D.", "Flavor Flav", "Professor Griff", "Khari Winn", "DJ Lord"],
["Axl Rose", "Slash", "Duff McKagan", "Steven Adler"],
["Run", "DMC", "Jam Master Jay"],
]
# Your code here
def test(params):
for param in params:
print(*param, sep = ", ")
test(musical_groups)
2 Answers
![Steven Parker](https://uploads.teamtreehouse.com/production/profile-photos/969492/micro_tool_chess_set_2.jpg)
Steven Parker
230,688 PointsSome challenges are as much about practicing a specific method as they are about producing the expected output. The instructions in this case asked you to "output the members joined together with a ", " comma space as a separator".
Your approach would produce the same output but it doesn't join the members together (using the "join" string method).
![Robson Possidonio](https://secure.gravatar.com/avatar/1c237572f79289d1d79fdd1982188d6f?s=96&d=https%3A%2F%2Fecs-static.teamtreehouse.com%2Fassets%2Fcontent%2Fdefault_avatar-445fbbabfc8dc9188fb5967fe43322ee0c3e0dd1e10f378bf8343784af5a13eb.webp&r=pg)
Robson Possidonio
977 Pointsunderstood. thanks!