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?

SmartMathMan Jan 9, 2018

#1**+1 **

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

So

4 * 3 * 2 * 1 = 24 ways

CPhill Jan 9, 2018

#2**0 **

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.

SmartMathMan Jan 9, 2018

#3**+1 **

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 :

n^{k} = 4^{4} = 256 ways

Where k = the balls [ the four bracelets ] and n = the boxes [ the four people]

CPhill Jan 10, 2018