(a) How many cards are in a complete deck of Set? (b) How many unique sets are there? When three cards form a set, we can also count the number of attributes for which all three cards are the same. For example, in the set below, all three cards have the same color, and the same shape (but they have different numbers of shapes and different shading). [asy] import patterns; add("hatch",hatch()); add("hatchred",hatch(1mm,W,red)); add("hatchpurple",hatch(1mm,W,purple)); add("hatchgreen",hatch(1mm,W,green)); unitsize(1 cm); real cardheight = 1.5, cardwidth = 2.5, cardspace = 0.2; int i, j; path card = (0,0)--(cardwidth,0)--(cardwidth,cardheight)--(0,cardheight)--cycle; path oval = arc((0,0.25),0.2,0,180)--arc((0,-0.25),0.2,180,360)--cycle; path diamond = (0.2,0)--(0,0.5)--(-0.2,0)--(0,-0.5)--cycle; path squiggle = (0.25,0.2){dir(90)}::(0,0.5){dir(160)}::(-0.2,0.4){dir(250)}::(-0.1,0.15){dir(270)}::(-0.25,-0.2){dir(270)}::(0,-0.5){dir(340)}::(0.2,-0.4){dir(70)}::(0.1,-0.15){dir(90)}::cy

Guest Jan 29, 2022

#1**0 **

(b) There are 108/3 = 36 unique sets.

(c) For the first card, there are 3 ways to choose the number. For the second, there 2 choices, and then there is only 1 choice. So there are 3*2*1 = 6 ways that the numbers can be chosen. Doing this for the other attributes, we get 6*6*24*6 ways. But the order of cards doesn't matter in a set, so we divide by 3!: 6*6*24*6/3! = 864. So there are 864 sets for part (c).

(d) The cards can have the same number, color, shape, or shading. If all the colors are the same, then there are 6*24*6/3! = 144 sets. If all the numbers are the same, then there are 144 sets. We get the same number for shape and shading, so there are 4*144 = 576 sets for part (d).

(e) We need to choose two of the attributes. There are C(4,2) = 6 ways of choosing two attributes. For each of these two attributes, there are 3 options. For the other two attributes, there are 3 ways of assigning the choices, so there are 6*3*3*3*3 = 486 sets for part (e)

(f) First we choose which attributes are the same. There are C(4,3) = 4 ways of choosing which attributes are the same. There are then 3*3*4 = 36 ways to assign which is which for each of these three attributes, and there are 4 ways to assign the choices for the fourth attribute, so there are 4*36*4 = 576 sets for part (f).

Guest Mar 6, 2022