In how many ways can the 2 x 10 board shown below be covered with 1 x 2 dominos?
Hello, BKelAB!
I think I have an approach,
. . but it requires a bit of Listing.
There are two types of dominos:
. . Horizontal $\displaystyle (H)\!:\;\;\sqsubset \!\sqsupset $
. . . . Vertical $\displaystyle (V)\!:\;\;\begin{array}{c} \sqcap \\ [-3mm] \sqcup \end{array}$
Note that two $\displaystyle H$'s form a 2-by-2 square.
. . Let $\displaystyle H\!H$ represent: .$\displaystyle \begin{array}{c}\sqsubset\! \sqsupset \\ [-3mm] \sqsubset\! \sqsupset \end{array}$
There are 6 cases to consider:
(1) 10 V's: there is 1 way.
(2) 8 V's, 1 HH
. . .Place the 8 V's in a row, leaving spaces before, after and between them.
. . . . $\displaystyle \_ \;V\;\_\;V\;\_\;V\;\_\;V\;\_\;V\;\_\; V\;\_\;V\;\_\;V\;\_ $
. . .The $\displaystyle H\!H$ can be placed in any of the 9 spaces.
. . There are 9 ways with 8 V's and 1 HH.
(3) 6 V's, 2 HH's
. . .Place the 6 V's in a row, leaving spaces before, after and between them.
. . . . $\displaystyle \_\;V\;\_\;V\;\_\;V\;\_\:V\;\_\;V\;\_\;V\;\_$
. . .Each HH has 7 choices of spaces: .$\displaystyle 7^2$ choices.
. . .There are 49 ways with 6 V's and 2 HH's.
(4) 4 V's, 3 HH's
. . .Place the 4 V's in a row, leaving spaces before, after and between them.
. . . . $\displaystyle \_\;V\;\_\;V\;\_\;V\;\_\;V\;\_$
. . .Each HH has 5 choices of spaces: .$\displaystyle 5^3$ choices.
. . .There are 125 ways with 4 V's and 3 HH's.
(5) 2V's, 4HH's
. . .Place the 4 HH's in a row, leaving spaces before, after and between them.
. . . . $\displaystyle \_\;H\!H\;\_\;H\!H\;\_\;H\!H\;\_\;H\!H \;\_$
. . .Each V has 5 choices for spaces.: .$\displaystyle 5^2$ choices.
. . .There are 25 ways with 2V's and 4 HH's.
(6) 5 HH's: There is 1 way.
Therefore, there are: .$\displaystyle 1 + 9 + 49 + 125 + 25 + 1 \;=\;{\color{blue}210}$ ways.