A
Quad-Omino (not to be confused with a
tetromino) is a square-shaped tile with numbers in each corner. Like the game of dominoes, the object is to place them next to each other so their numbers match.
The standard set I own with numbers 0 to 5 has 125 tiles. Does the set really contain every possible tile?
Find a formula for the total number of different tetrominoes in a set numbered from 0 to n.
1) How many possibilities are there for the numbers on a single tile? This is "6 choose 4" or {6 4} = 6!/(4! 2!) = 15. More generally, for Quad-Ominos, numbered 0 to n, there would be {(n+1) 4} = (n+1)!/(4! (n-3)!) = (1/24) (n+1)! / (n-3)! different sets of 4.
Even more generally, for m-Ominos, numbered 0,1, ..., n, there would be {n+1 m} sets of m.
2) Now, how many arrangements of each set of 4 numbers are possible? (I will not call these cyclic permutations, since this term technically has a different meaning.) Choose an arbitrary element. One of 3 remaining elements can be placed going around CW, and then one of 2 remaining elements can be placed CW to that. So there are 2x3=6 arrangements. So Jer's set would need 15 * 6 = 90 tiles at a minimum to be complete. So, with 125, he likely has a complete set and then some.
For any four numbers on a tile, the number of arrangements is still 6, so a set of Quad-Ominos going 0 to n needs to have (6/24).... or
(1/4) (n+1)!/(n-3)! tiles.
For m-Ominos. the number of arrangements of m numbers is (m-1)(m-2)...(1), or (m-1)! So, a complete set of m-Ominos, with digits 0 to n, would have a minimum of (m-1)! {n+1 m} tiles.
Edited on October 20, 2021, 9:57 am
possible solution
