In an $8 \times 2$ grid, you can place $5$ blue pins, $2$ red pins and $1$ yellow pin. The only rule is that the two red pins cannot share an edge. How many different arrangements of the pins can be made?
Looking at the arrangements top to bottom are considered different arrangements, as are using left and right columns.
I think you can have 36 different arrangements for the red pins and 8 for the yellow pin, and since the blue pins are all the same, does this mean that the answer would be 288? I feel like my reasoning has gone wrong somewhere though, so if you could offer any advice, thanks.
Considering all rotation as different arrangements (as clarified).
$2$ red pins, $5$ blue pins, $1$ yellow pin in $8$ x $2$ grid. Cells with red pins cannot share an edge (note, they can share a corner).
Say $8$ x $2$ grid is $2$ columns and $8$ rows. You put one red pin in the left top cell. Then you have $13$ valid cells to place the second red. Similarly for the left top cell. That makes it $13 \times 2$. Now let's come to the second row from top. Please note we have to ignore the first row as any possible (which is only $1$ each in this case for left and right) combinations is already counted. Now you have $2 \times 11$. Similarly keep striking off rows that are done and this gives you
$2 \times (13 + 11 + 9 + 7 + 5 + 3 + 1) = 98$ possibilities for red pins.
This leaves $14$ cells. You can choose $5$ cells for blue pins from it in ${14 \choose 5}$ ways. You are then left with $9$ cells and you can place $1$ yellow pin in $9$ ways.
That gives total of $98 \times {14 \choose 5} \times 9 = 1765764$ arrangements.