From Curtis Karnow comes a puzzle that boldly goes where no human has gone before:
You are creating a variation of a Romulan pixmit deck. Each card is an equilateral triangle, with one of the digits 0 through 9 (written in Romulan, of course) at the base of each side of the card. No number appears more than once on each card. Furthermore, every card in the deck is unique, meaning no card can be rotated so that it matches (i.e., can be superimposed on) any other card.
What is the greatest number of cards your pixmit deck can have?
Extra credit: Suppose you allow numbers to appear two or three times on a given card. Once again, no card can be rotated so that it matches any other card. Now what is the greatest number of cards your pixmit deck can have?
Three numbers in a row can be calculated by applying $n(n-1)(n-2)$ for $n$ numbers/symbols to be used. Dividing by three accounts for the three different rotational orientations of an equilateral triangle.
Allowing numbers to appear two times on a card means that if that occurs, the card has exactly two numbers. This can occur $n \cdot (n-1)$ ways, as any number can appear twice, and any other number can appear once.
Allowing numbers to appear three times on a card means that if that occurs, the card has exactly one number. This can occur $n$ ways, as any number can be the number that appears 3 times.
The following table summarizes the results:
| # of Numbers/Symbols on Cards | Numbers Appear Exactly Once | One Number Appears Exactly Twice | One Number Appears Exactly Thrice |
|---|---|---|---|
| $0$ | $0$ | $0$ | $0$ |
| $1$ | $0$ | $0$ | $1$ |
| $2$ | $0$ | $2$ | $2$ |
| $3$ | $2$ | $6$ | $3$ |
| $...$ | $...$ | $...$ | $...$ |
| $n$ | $\dfrac{n(n-1)(n-2)}{3}$ | $n(n-1)$ | $n$ |
Substituting $n = 10$ yields $\dfrac{10 \cdot 9 \cdot 8}{3} = 240$ unique cards.
Allowing repeats,
$$\dfrac{n(n-1)(n-2)}{3} + n(n-1) + n$$
$$= \dfrac{n(n-1)(n-2)}{3} + \dfrac{3n(n-1)}{3} + \dfrac{3n}{3}$$
$$= \dfrac{n(n-1)(n-2+3)}{3} + \dfrac{3n}{3}$$
$$= \dfrac{n(n^2-1+3)}{3}$$
$$= \dfrac{n^3+2n}{3}$$
Substituting $n = 10$ yields $\dfrac{1020}{3} = 340$ unique cards, allowing repeatable numbers.