April 18, 2021
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?
Answer: 240 cards for original puzzle; 340 for Extra Credit puzzle.
Explanation:
We are going to apply the "choose" function in creative ways to solve the puzzle and the Extra Credit. Simply put, the "choose" function is mathematically defined as the number of ways to choose \(k\) items out of \(N\) items, for \(0 \leq k \leq N\), written as \(\binom{N}{k} = \frac{N!}{k!(N-k)!}\). For the original problem, since no number appears more than once on each card, it means that we have to choose, quite literally, 3 digits out of the possible 10 digits from 0 to 9. This means here \(N = 10\) and \(k = 3\). Therefore, we must involve the value \(\binom{10}{3} = 120\) in our answer. From the uniqueness proposition of the puzzle, once we've chosen our 3 digits, we must consider how many unique configurations there are. The answer is that there are two unique configurations for each trio of digits. To see this, consider the digits "0", "1", and "2". Here either "1" is to the left of "0" or to the right. No amount of rotation can bring about one configuration to the other (and flipping the card is not allowed!). The third digit, "2", of course, is fixed once the relative position of "1" with respect to "0" is decided. Therefore, the answer to our original puzzle is simply \(2 \binom{10}{3} = 2 \times 120 = \boxed{240}\) cards.
For the Extra Credit, we consider the total number of cards possible by breaking down into three cases in terms of number of distinct digits appearing on the cards. The cases are:
So we only need to find out the case where two of the sides are the same digit and the third one is distinct. Here once again we have 10 digits to choose from and so our \(N = 10\) as before. For \(k\), since we only need to consider picking out two digits, \(k = 2\). Therefore, we need to involve the value \(\binom{10}{2} = 45\). Here, there is no longer uniqueness in the relative positions of the digits as two of them are the same, but rather a decision needs to be made on which of the two digits should appear twice on the card. Since there are two choices, for each pair of digits chosen, we can have two unique configurations once again. Therefore, the total cases for two distinct digits is \(2\binom{10}{2} = 90\).
Putting everything together, the answer to the Extra Credit is \(240 + 90 + 10 = \boxed{340}\) cards.