+0  
 
+1
571
3
avatar+717 

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
avatar+91099 
+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

 

 

cool cool cool 

CPhill  Jan 9, 2018
 #2
avatar+717 
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
avatar+91099 
+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 : 

 

nk     =  44  =    256 ways  

 

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

 

 

cool cool cool

CPhill  Jan 10, 2018

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