Take the 15 smallest dominoes in a set (double blank through double four.)

In how many ways can they be arranged in a row such that the numbers on consecutive pieces match.

Count the two directions separately.

I had the same simplification in mind as Josh, but I calculated it differently.  Same answer, though: Solve the problem without doubles, and then multiply by 48.

My thinking:
a) Instead of thinking in terms of strings, think in terms of complete circles.  I expect that this will simplify the problem, because it avoids the problem of what is happenning at the ends.   Join the start of the row to the end in order to form a circle.

b) What ever the number of unique 10 domino rows, divide by 10 to get the number of unique 10-domino circles.  This is because any 10-domino circle can be broken in any of ten points to form a unique row.

c) Any double can be inserted into the circle in one of two ways.  Since there are 5 doubles, all doubles can be inserted in
2x2X2X2X2 = 32 different ways, forming a 15-domino circle.

d) Each 15 domino circle can be converted into 15 unique rows, by breaking it at any one of 15 points.
 
So, whatever the 10-domino answer is, multiply it by 32*15/10 = 48 to get the 15-domino answer.

Posted by Steve Herman on 2006-05-28 08:45:32