How many ways are there to divide a red bracelet, a yellow bracelet, a green bracelet, and a black bracelet among 4 people if each person can receive any number of bracelets, including ? Each bracelet must be given to someone.
+36916 Does it say that each must be given at least one bracelet? Not sure....
Here is my 'stab' at a solution....(I almost never get these correct BTW 
)
For EACH person:
4 bracelets 1 way
3 Bracelets 4 ways
2 bracelets 6 ways
1 bracelet 4 ways
0 bracelet 1 way Total 16 ways x 4 people = 64 possible ways (??) Some of these may be 'reflections' of the others....so .... I really do not KNOW !
+128460 Each bracelet must be given to someone, but....I believe that the person who posted the question probably meant that each person can receive any number of bracelets including no bracelets at all
+6248 We'll go with each person gets 0-4 bracelets.
This is just the problem of 4 distinct balls into 4 distinct bins.
Each bracelet can just be tagged with with person it is allotted to i.e.
each arrangement is a 4 digit base 4 number so there are
\(n = 4^4 = 256 \text{ arrangements}\)
+36916 Hmmmmm..... I still don't get this.....
example base 4 number 3 3 3 3 is 12 bracelets .... there is only 4 total ...plus the possibilty of 0.
so I think you would have to eliminate any base 4 number that totals more than 4 .... Yes? No?
+4609 Is this question even complete? How many ways are there to divide a red bracelet, a yellow bracelet, a green bracelet, and a black bracelet among 4 people if each person can receive any number of bracelets, including ? Each bracelet must be given to someone. Including what?
A quick search gives web2.0calc link to the same question. Look at part (b). The full question is, "How many ways are there to divide a red bracelet, a yellow bracelet, a green bracelet, and a black bracelet among 4 people if each person can receive any number of bracelets, including 0? Each bracelet must be given to someone."
CPhill correctly answered the question by using distinguishability, where it is \(n^k\) with duplicates and \(k^n\) without duplicates. This does require duplicates, CPhill is correct.
