so i was thinking stars and bars, but the people arent distinguishable and the bracelets are ....
So casework is the only option.
Complementary casework has less cases, so lets jump in
well.. we know that there can be 4^4 = 256 total cases
3 bracelets: choose 1 person to receive the 3 and one more out of the 3 remaining to get teh rest
and since distinguishable we have 3! * 1 * 3 = 18
4 bracelets: choose 1 person out of 4: 4 * 4! = 96
so 256-(18+96).... im not so sure ...
The link provided below has a correct answer, but the answerer didn't explain why it only works if it is divisible by 3.
First off, if you have a fraction n/2, then you can have any n in order to repeat
If you have a fraction n/3, then we need a multiple of 3
if we have a fraction n/5, then you can have any n
so we need a multiple of 3 in order to repeat
thus, the answer is 474/3 (if its divisible) or ceil(474/3) (if it's not divisible.)
it's divisible, so the answer is 474/3=158
There is a reason for the pattern - can you see why? Do you see anything intesting if we generalize \(x\) for not just 2, but for all integral \(x\)? Try finding a general formula. Also notice - does your 1, 0, etc. pattern have anything to do with \(x=2\)? Maybe there is, and maybe not. Try to find out. :)