+65 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.
+6248 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.
+23246 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.
