+0

0
430
5

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
+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
+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
+1

Thanks, Rom and Geno......!!!!   CPhill  May 11, 2019
#4
+1

It is reallzy good to see you one the forum Geno Melody  May 12, 2019
#5
+1

Thanks!

May 18, 2019