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 believe I can simplify this....
Ok arrange all the dominoes except the doubles in the manner described by the problem.
There are now 4 doubles that could fit in 2 spaces and 1 double that could fit in 3.
Ok, place those 4 doubles in any spots you like and consider moving different numbers of them. If you move none there is 1 possible arrangement of doubles, if you move 1 there are 4 possible arrangements of doubles, if you move 2 there are 6 possible arrangements of doubles, if you move 3 there are 4 possible arrangements of doubles, and if you move all 4 there is 1 possible arrangement of doubles.
This makes 16 possible arrangements for the 4 doubles. Each of these could have the last double in 3 different spaces. So, there are 16x3=48 total arrangements for the doubles. Now you just have to figure out the number of combinations for the ten non-double dominoes and multiply by 48. Hope this helps...
I think my reasoning is correct, but it is 8:00 a.m. and I did not sleep last night...
|
|
Posted by Josh
on 2006-05-28 08:30:54 |
No Subject
