+1438 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?
+128407 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
+1438 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.
+128407 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]
