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In the card game Set, each card features a number of shapes, with four attributes:

Number: The number of shapes is 1, 2, or 3.
Color: Each shape is red, purple, or green.
Shape: Each shape is oval, diamond, or squiggle.

There is exactly one card for each possible combination of attributes.

In the game, several of the cards are dealt out, and the goal is to find a set. A set is formed by three cards, where for each attribute, either all three cards are the same, or all three cards are different. When three cards form a set, we can also count the number of attributes for which all three cards are the same.

(a) How many cards are in a complete deck of Set?

(b) How many unique sets are there?

(c) Find the number of sets where all three cards are the same for exactly 0 attributes.

(d) Find the number of sets where all three cards are the same for exactly 1 attribute.

(e) Find the number of sets where all three cards are the same for exactly 2 attributes.

(f) Find the number of sets where all three cards are the same for exactly 3 attributes.

Jul 13, 2022

#1
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(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).

Jul 13, 2022
#2
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I believe you are wrong, for I worked out part b. There are 1080 unique sets.

Guest Jul 13, 2022
edited by Guest  Jul 13, 2022
#3
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Here is my attempt :

a)  3^4 = 81 cards

b)  81C3 = 85320 sets of 3 cards

c)    All different       3! *3!*3!  =  216

d)  Say they are all red  and the other attributes are all different  3!*3!*3!   = 216

But there are 12 different things that could be the same so that is  12*216 = 2592

e)    exactly 2 matching atributes

Say they are all R1  they are each allotted a shape.  then there is 3! choices for shading.

12C2 *3!  = 66*6 = 396

f)   say they are all the same except one is red, one Purple, and one green.  There is one set of those.

So there is one set  for each attribute.  so that is 4

Jul 18, 2022
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Melody  Jul 18, 2022
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Oct 7, 2022
#6
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Solution:

(a) There are four attributes, and there are three choices for each attribute. In a Set deck, there is one card for each combination of choices, so there are $$3^4=\boxed{81}$$ cards.

(b) Note that if we choose any two cards, there is exactly one card that will complete a set. For example, consider the following two cards: Since the first card has three shapes and the second card has one shape, the third card must have two shapes. Since both cards are red, the third card must also be red, and so on. Each attribute of the third card is determined uniquely by the first two cards, which gives us the following set: There are $$\dbinom{81}{2}$$ ways to choose two cards, and two cards uniquely generate a set. However, for each set, there are 3 ways that we could have chosen the two cards that generate the set. Therefore, the total number of possible sets is $$\dfrac{1}{3}\dbinom{81}{2}=\boxed{1080}.$$

(c) We want all three cards to be different with respect to every attribute.

There are $$81$$ ways to choose the first card in the set. Then for the second card, there are 2 ways to choose each attribute (since every attribute must be different from the first card), so there are $$2^4=16$$ ways to choose the second card. Then the third card is uniquely determined (since every attribute must be different from the first two cards).

This gives us a count of $$81\cdot16=1296.$$ But the order of the cards does not matter, so there are $$\dfrac{1296}{3!}=\boxed{216}$$ sets in this case.

(d) There is one attribute where all three cards are the same, and they are different for every other attribute.

There are 4 ways to choose the attribute for which the cards are the same, then 3 options for this attribute. There are $$3^3=27$$ ways to choose the first card (one attribute has already been determined), then $$2^3=8$$ ways to choose the second card, and then the third card is uniquely determined.

So there are $$\dfrac{4\cdot3\cdot27\cdot8}{3!}=\boxed{432}$$ sets in this case.

(e) There are two attributes where all three cards are the same, and they are different for every other attribute.

There are $$\dbinom{4}{2}=6$$ ways to choose the two attributes for which they are the same, then 3 options for these attributes. There are $$3^2=9$$ ways to choose the first card, then $$2^2=4$$ ways to choose the second card, and then the third card is uniquely determined.

So there are $$\dfrac{6\cdot3^2\cdot9\cdot4}{3!}=\boxed{324}$$ sets in this case.

(f) There are three attributes where all three cards are the same, and they are different for the remaining attribute.

There are $$\dbinom{4}{3}=4$$ ways to choose the three attributes for which they are the same, then 3 options for these attributes. There are exactly three cards that meet these conditions, which form a set.

So there are $$4\cdot3^3=\boxed{108}$$ sets in this case.

As a check, the answers for parts (c)-(f) add up to $$216+432+324+108=1080$$ which is the answer for part (b).

Oct 8, 2022