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0
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avatar+64 

Part A ~

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?

 

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.

 May 11, 2019
 #1
avatar+6196 
+2

A is trivial.  This sort of problem is one of the first things you learn in discrete math class.  Think permutations.

 

B is less so. 

 

Each bracelet can be tagged with a base 4 digit indicating to which person it's assigned.

There are 4 bracelets so the different assignments can be represented by a 4 digit base 4 number.

 

There are 44= 256 of these numbers and hence 256 different assignments.

 May 11, 2019
 #2
avatar+21953 
+4

For Part B, same answer but a different approach --

 

Part A:  The first person can receive any of 4 colors, the second person can receive any of 3 colors, the third person can receive any of 2 colors, while the last person can receive only the color remaining     --->     4 x 3 x 2 x 1  =  24

 

Part B: Call the 4 bracelets A, B, C, and D; '0' means that a person receives no bracelet.

 

Case 1: each person receives exactly one bracelet -- (see Part A) --  24 ways

 

Case 2: one person receives all four bracelets -- since any of the 4 persons can get them all, there will be 4 ways:

                (A, B, C, D) - 0 - 0 - 0     or     0 - (A, B, C, D) - 0 - 0     or     0 - 0 - (A, B, C, D) - 0     or     0 - 0 - 0 - (A, B, C, D)

 

Case 3: one person receives three of the bracelets, while one of the other three persons gets the other one:

                (A, B, C) - D - 0 - 0     or     D - (A, B, C) - 0 - 0     or     D - 0 - (A, B, C) - 0     or     D - 0 - 0 - (A, B, C)     4 ways

                (A, B, C) - 0 - D - 0     or     0 - (A, B, C) - D - 0     or     0 - D - (A, B, C) - 0     or     0 - 0 - D - (A, B, C)     4 ways

                (A, B, C) - 0 - 0 - D     or     0 - (A, B, C) - 0 - D     or     0 - 0 - (A, B, C) - D     or     0 - 0 - D - (A, B, C)     4 ways

                                                                                                                                                              subtotal   =   12 ways      

 

But instead of (A, B, C), the three could be (A, B, D), (A, C, D), or  (B, C, D), so  4 x 12  =  48 ways

 

Case 4: one person receives two bracelets, two other persons each receive one bracelet, and one person does not get a bracelet:

                 (A, B) - C - D - 0     or     (A, B) - C - 0 - D     or     (A, B) - D - C - 0     or     (A, B) - D - 0 - C

                         or     (A, B) - 0 - C - D     or     (A, B) - 0 - D - C            subtotal  =  6 ways

              The set of (A, B) could be in each of 4 different positions      subtotal  =  4 x 6  =  24 ways

              Instead of (A, B),  the set of two could be (A, C), (A, D), (B, C), (B, D), or (C, D)      =  6 ways

              Total  =  6 x 4 x 6  =  144 ways

 

Case 5:  two persons each get two bracelets, while the ohter two persons get no bracelets:

                  (A, B) - (C, D) - 0 - 0     or     (A, B) - 0 - (C, D) - 0     or     (A, B) - 0 - 0 - (C, D)

                  (C, D) - (A, B) - 0 - 0     or     (C, D) - 0 - (A, B) - 0     or     (C, D) - 0 - 0 - (A, B)

                  0 - (A, B) - (C, D) - 0     or     0 - (A, B) - 0 - (C, D)     or     0 - 0 - (A, B) - (C, D)

                  0 - (C, D) - (A, B) - 0     or     0 - (C, D) - 0 - (A, B)     or     0 - 0 - (C, D) - (A, B)             subtotal  =  12 ways

               Instead of splitting the bracelets as (A, B) and (C, D), they could be split as (A, C) and (B, D) or as (A, D) and (B, C)

               Total  =  3 x 12  =  36 ways

 

Final amount:  24 + 4 + 144 + 48 + 36  =  256 ways.

 May 11, 2019
 #3
avatar+111456 
+1

Thanks, Rom and Geno......!!!!

 

 

cool cool cool

CPhill  May 11, 2019
 #4
avatar+110715 
+1

It is reallzy good to see you one the forum Geno   laugh

Melody  May 12, 2019
 #5
avatar+64 
+1

Thanks!

 May 18, 2019

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