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 at most two bracelets? Each bracelet must be given to someone.

Guest Jan 29, 2022

#1**+1 **

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 ...

MathProblemSolver101 Jan 29, 2022

#2**+1 **

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 at most two bracelets? Each bracelet must be given to someone.

1 each 4! ways = 24 ways

1 pair two individuals 4C2=6 4*3*2*6 = 144 ways

2 pairs 4C2/2 = 3 4*3 *3 = 24 ways

24+24+144 = 192 ways

I think.

Melody Jan 29, 2022