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!