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.
Thanks, Charlie. I agree with all of your numbers, and your final
answer. I did miss one possibility in the 3 and 7 string
sub-case, but my other errors were just multiplication (since I did
correctly include reversals). Once, I multiplied 4x3x2 and got
12! Once, I multiplied 4x3 and got 6! A good argument for a
computer, or at least a calculator! :-)
So here is my corrected (and fuller solution):
a) Consider complete 10-chains without doubles. (See my previous
post). Cut these complete chains in the two places the zeroes
join. This
forms either a 3-row and a 7 row, or a 4 row and a 6 row, or two 5 rows.
b) The 5 row must involve 1 and 2 and 3 and 4, in some order, so there
are 4x3x2x1 = 24 different 5-rows. Each 5 row could have two
different second matching 5-rows (using the remaining dominoes), so it
looks like there are 48 chains. But this double-counts the case
where the first 5-row is the second one, so dividing by 2 yields 24
unique chains that yield a pair of
five rows when broken at each of the zeros.
c) The 4 row must involve 3 different non-zero numbers, in some order,
so there are 4x3x2 = 24 different 4 rows. The matching 6 row has
two choices for which end to put the domino with 0 and the number not
used in the 3 row, and two choices on which domino to attach to it, for
a total of 4 unique matching 6-rows using the dominos not in the 4
row. Altogether, 24 x 4 = 96 unique chains that yield a pair of
4-row and a 6-row when broken at each of the zeros.
d) The 3 row must involve 2 different non-zero numbers, in some
order, so
there are 4x3 = 12 different 3 rows. The matching 7 row has two
choices for which way to put the two remaining domino with 0, and then
6 different ways to put in the remaining 5 dominos, for atotal of 12
unique matching 7-rows using the dominos not in the 3 row.
Altogether, 12 x12 = 144 unique chains that yield a 3-row and a 7-row
when broken at each of the zeros.
e) Altogether, 24 + 96 + 144 = 264
different unique 10-domino chains (without doubles).
f) Doubles can be inserted in 2x2x2x2x2 = 32 different ways, giving 8488 unique 15-chains, including doubles.
g) And each of these 15 chains can be cut in 15 ways, with each cut
yielding a unique row. Final (and correct, thanks to Charlie) answer
= 8488 * 15 = 126,720 15-domino rows where consecutive pieces
match.
And I agree with Charlie
