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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.

 Dec 17, 2020
 #1
avatar+419 
+1

Case 1: 0, 0, 2, 2 and the combinations.

 

There are 6 ways to arrange 0,0,2,2 and 1 way for first and second, 6 for the last two, so 36 * 6 = 216.

Case 2: 0,1,1,2 and the combinations

 

There are 12 ways to arrange 0,1,1,2, and 1 way for the first person, 4 ways for second, 3 ways for third, and 2 ways for fourth, so 24 * 12 = 288.

Case 3: 1,1,1,1

 

4 * 3  * 2 *. 1 = 24

 

216 + 288 + 24 = 240 + 288 = 328? I'm probably wrong, this is my best attempt.

 Dec 17, 2020
 #2
avatar+112865 
+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.

 

Let's take a look Pangolin14

 

I am assuming that the people are all different.

 

division of bracelets.

4,0,0,0      4 ways

3,1,0,0      4 ways to chose odd one out, 4 ways to choose who to give the 3 ways to choose who gets the one.  

                  4*4*3 = 48 ways

 

2,2,0,0      4C2 /2  =  3 ways to pair the bracelets

                 4 ways to chose the allocate the fist pair,  3 ways to allocate the 2nd pair

                  3*4*3 = 36 ways

 

2,1,1,0      4C2 = 6 ways to choose the pair.

                 4! ways to order the people = 24 ways

                  6*24 = 144 ways

 

1,1,1,1      4! ways = 24 ways

 

4+48+36+144+24 = 256 ways

 

I am happy to discuss this if you want guest or Pangolin14  laugh

 Dec 17, 2020
 #3
avatar+419 
+1

I seem to have overlooked the 4,0,0,0 case... 

 

I might be wrong, but for the 2,1,1,0 case the number of ways should be the number of arrangements of 2,1,1,0 * number of ways to pick, and the number of arrangements is 4!/2! = 12, and the number of ways to pick the bracelets: 4C2 for the first pair, 2C1 for the second, 1C1, multiply, 6 * 2 = 12, then * 24 = 288, so perhaps the pair of 1s was left out in your solution?

Pangolin14  Dec 18, 2020
 #4
avatar+112865 
+1

Thanks Pangolin14,

I am really pleased that you have interacted with me.   laugh

 

I am just going to look at the  2,1,1,0 arrangement

 

I do not understand how you are looking at the problem.

I see 12  arrangements of bracelets and 12 arrangements of people =144

but then for some reason, you double your answer ... I do not understand why.

 

Here is my logic,

 

FIRST assume all the bracelets are the same.

I am going to order the people so that the first person in the queue gets 2 bracelets, the next two people get 1 bracelet each and the last person gets no bracelets.

There are 4! ways to order 4 people.  but it does not matter what order the middle 2 are in because they are both going to get 1 bracelet anyway so I have to divide by 2.

4!/2 = 12 ways

 

Now how many ways can I queue the bracelets?  that would also be 4!

BUT it does not matter what order the first two are in because they are going to get given to the same person.

so that is

4!/2 = 12 ways

 

So I am reasonably confident that the answer is 12*12 = 144 ways

Melody  Dec 19, 2020
edited by Melody  Dec 19, 2020
 #5
avatar+419 
+1

Oh, calculation error. In the previous post I basically meant what you had written, except somehow multiplied by 24 instead of 12...?

Pangolin14  Dec 20, 2020

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