Find the number of *sets* where all three card are the same for exactly $0,1,2,3$ attributes. 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.
Shading: Each shape is hollow, shaded, or striped.
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.

Problem:
(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.

My attempt
(a) $3^4=\boxed{81}$. Since there are 4 different attributes and 3 different choices for each one. $3\cdot3\cdot3\cdot3=81$
(b) $\dfrac{C(81,2)}{3}=\boxed{1080}$. We know 81 is the total number of cards(part a). We also know that there is exactly one out of any two cards that will make a set, so you get $\dbinom{81}{2}$, which is 3240. Then, you have to divide by 3 because you overcounted, so the answer is $\dfrac{3240}{3}=\boxed{1080}$
(c) $\dfrac{81\cdot16\cdot1}{6}=\boxed{216}$.We know that there could only be 81 ways to choose the first one, while there could only be $2^4=16$ cards for the second one because you already chose one of each attribute, which means 2 things you can choose for each attribute. A similar thing happens with the last card, which is $1^4=1$. Then,  you multiply them:$81\cdot16\cdot1$. After that, you have to make up for overcounting, so you divide $81\cdot16\cdot1$ by $3!=6$, and you get $\boxed{216}$.
(over here I am not sure about the "$2\cdot2^4=32$" part)
(d) $\dfrac{81\cdot32\cdot1}{6}=\boxed{432}$. Again, like part c, we start with 81 ways. The second one is different. Since this time the cards could be the same for one attribute, then there are $2\cdot2^4=32$ combinations. Then, the last one has only one way. Again we make up for overcounting by $3!$. So after you multiply $81\cdot32\cdot1$ and divide that by $6$ you get 432
For the (e) and (f) parts, I have no inkling about what to do. Please help!
 A: We can find the number of Sets (as the actual game's instruction manual styles them) with $k$ common attributes for all $0\le k\le3$ in one fell swoop. Fixing $k$, the count is the product of:

*

*$\binom4k$ ways to select which attributes are common

*$3^k$ ways to specify the nature of these common attributes (e.g. for shape, "oval", "diamond" and "squiggle" are the possible natures)

*$(3!)^{4-k-1}=6^{3-k}$ ways to specify the natures of the remaining, distinct attributes. The order of cards in a Set can be canonicalised by sorting on the natures of the earliest non-common, i.e. distinct attribute. The natures may then be assigned to the cards in arbitrary order – $3!$ ways for each distinct attribute and $3-k$ free distinct attributes gives this factor.

Evaluating $\binom4k3^k6^{3-k}$ for $k=0,1,2,3$ gives the answers for the last four parts as $216,432,324,108$ respectively, for a total of $1080$ Sets.

Reformulated in the method you used for (c) and (d):

*

*The first card may be any one of the $81$ cards of the game

*For the second card, choose which attributes are the same and which are different. You then have $2$ choices for a "different" attribute and $1$ for a "same" attribute, so $\binom4k2^{4-k}$

*The third card is uniquely determined from the first two if you want to form a Set

*Divide by $6$ for overcounting

A: 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 $\binom{81}{2}$ 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 $\frac{1}{3} \binom{81}{2} = \boxed{1080}.$
(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 $81 \cdot 16 = 1296.$ But the order of the cards does not matter, so there are $1296/3! = \boxed{216}$ 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 $4 \cdot 3 \cdot 27 \cdot 8/3! = \boxed{432}$ 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 $\binom{4}{2} = 6$ 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 $6 \cdot 3^2 \cdot 9 \cdot 4/3! = \boxed{324}$ 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 $\binom{4}{3} = 4$ 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 $4 \cdot 3^3 = \boxed{108}$ 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).
