Number of winning tern in a deck of cards and other 3 related questions There is a deck made of $81$ different card. On each card there are $4$ seeds and each seeds can have $3$ different colors, hence generating the $ 3\cdot3\cdot3\cdot3 = 81 $ card in the deck.
A tern is a winning one if,for every seed, the correspondent colors on the three card are or all the same or all different.
-1 How many winning tern there are in the deck?
-2 Shows that $4$ cards can have at most one winning tern.
-3 Draw $4$ card from the deck. What is the probability that there is a winning tern?
-4 We want delete $4$ cards from the decks to get rid of as many tern as possible. How we choose the $4$ cards?
I've no official solution for this problem and i don't know where to start.
Rather than a complete solution i would appreciate more some hints to give me the input to start thinking a solution.
 A: HINT: The problem is equivalent to the following one. Let $D=\{0,1,2\}^4$, the set of $4$-tuples of numbers from the set $\{0,1,2\}$. Each member of $D$ corresponds to a card; the four components of the $4$-tuple are the seeds; and $0,1$, and $2$ are the three ‘colors’. A tern is a set of three members of $D$.


*

*Prove that a tern $\{\langle a_1,a_2,a_3,a_4\rangle,\langle b_1,b_2,b_3,b_4\rangle,\langle c_1,c_2,c_3,c_4\rangle\}$ is winning if and only if $a_k+b_k+c_k\equiv 0\pmod3$ for $k=1,2,3,4$. If we think of the members of $D$ as vectors, then a tern $\{a,b,c\}$ is winning if and only if $a+b+c=\underline 0=\langle 0,0,0,0\rangle$, where the addition in each component is modulo $3$.


I think that this version is a little easier to work with.


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*To count the winning terns, show that if $a,b\in D$ and $a\ne b$, then there is exactly one $c\in D$ such that $a+b+c=\underline 0$. How many pairs of distinct elements can we choose from $D$? How many different pairs of distinct elements produce the same winning tern?

*This is a straightforward consequence of the first sentence in (1).

*Imagine drawing the $4$ cards one at a time. The first two can be anything. What’s the probability that the third doesn’t make a winning tern with the first two? Assuming that the first three cards aren’t a winning tern, what is the probability that the fourth does not complete a winning tern? It may be helpful to show that if $a,b$, and $c$ are distinct members of $D$, then the sums $a+b,a+c$, and $b+c$ are distinct as well.

*If $T_a$ is the set of winning terns containing the card $a\in D$, how many terns are in $T_a\cap T_b$ when $a\ne b$? Is $T_a\cup T_b\cup T_c$ bigger when $\{a,b,c\}$ is a winning tern, or when it’s a non-winning tern? Playing with these ideas should help you see how you need to choose the four deleted cards $a,b,c,d$ in order to make $T_a\cup T_b\cup T_c\cup T_d$ as large as possible.
A: For question one, ask a simpler question: Look at just the first seeds, how many ways can you color them?
Two possibilities are: Card 1: Red, Card 2: Red, Card 3: Red
and Card1:Red, Card2: Blue, Card3: Yellow.
Now for four seeds, you make the choice of colors four times.
You should check that you haven't chosen the same card three times - how many ways could you have done that?
