We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
158
5
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+5662 
+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+17776 
+3

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+102417 
+1

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

 

 

cool cool cool

CPhill  May 11, 2019
 #4
avatar+102792 
+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

3 Online Users