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.

There are 5 doubles but they can be rotated so there are 10 possibilities for the first selection.

Consider the just the ends of the other dominoes, there are four blanks, four 1's, four 2's, four 3's and four 4's.

For just one double, there are four possibilities for one end leaving 3 for the other; but the ends can be swapped. For that double we therefore have 12 * 2 for the end dominoes times 2 again for the rotation of the blank giving 24 thus far.

 [with 5 blanks that takes us up to 120, but I'm ahead of myself here. ]

Having placed 3 dominoes, those at the end can be interfaced 5 ways each, with a double (offers 2) and 3 others which match the end.

I haven't thought much beyond this, but I think I can see a possible 'algorithm' developing so that I wouldn't have to try to place all 15 dominoes.

I'll leave off here for the time.

Posted by brianjn on 2006-05-27 21:30:16