A problem involving a pack of $27$ distinct cards.

Question

Regional Mathematics Olympiad (India) $2019$ Leg $2$ Question $5$

There is a pack of $27$ distinct cards, and each card has three values on it. The first value is a shape from $\{\Delta,\square,\odot\}$; the second value is a letter from $\{A,B,C\}$; and the third value is a number from $\{1,2,3\}$. In how many ways can we choose an unordered set of 3 cards from the pack, so that no two of the chosen cards have two matching values. For example we can choose $\{\Delta A1,\Delta B2,\odot C3\}$ But we cannot choose $\{\Delta A1,\square B2,\Delta C1\}$

In this AoPS thread, two people got $36$ as the answer but I got $1278$ (I am Festus). Also, a solution here matches with my answer, though my solution was different.

I can't quote my answer verbatim, but here is a gist of my solution : I said that, if we calculate number of ordered sets and divide it by $6$, we shall get number of unordered sets. Then I proceeded to calculate by brute forcing(assuming cases), and arrived at $7668$, ordered sets and thus $1278$ unordered sets. Please let me know if I am correct or not.

Edit: as pointed out by Jose Maria, I made a typo in the example provided in the question, which has now been rectified. P.S. Copy-pasted it from the AoPS link.

Answer 1

Sorry. I can't comment due to reputation. The only way is to propose a solution. Check the allowed combinations in the two problems. Are differents!!

Here (solution 36) the example is {ΔA1,□B2,⊙C3}

In the other problem (solution 1278) the example is {ΔA1,ΔB2,⊙C3}

Answer 2

36 is the solution to the problem:

... pack, so that no two of the chosen cards have ANY matching values.

Your solution of 1278 is correct for the given problem, I would assume the solutions in AOPS have misread the problem, as I did at first as well.