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 must receive exactly 1 of the bracelets?
Any of the 4 could receive the red bracelet
After this.....any of the remaining 3 people without a bracelet could receive the yellow bracelet
After this.....either of the remaining 2 people without a bracelet could receive the green bracelet
And, by default, the remianing person without a bracelet will get the black bracelet
4 * 3 * 2 * 1 = 24 ways
this is part b
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.
The second question is analogous to the number of ways of putting 4 distinguishable balls into 4 distinguishable boxes with no restrictions [ that is....a box may contain 0 - 4 balls ]
The number of ways for this to happen is given by :
nk = 44 = 256 ways
Where k = the balls [ the four bracelets ] and n = the boxes [ the four people]