We have $10$ bricks, $3$ red, $2$ white, $2$ yellow, $2$ blue, $1$ black. In how many ways can these be arranged such that only $2$ red bricks are adjacent ?
We want to distribute the elements in {$RR, R$} over the spaces in arrangements of the form:
_B_B_B_B_B_B_B_
where _ represents a space (of which there are $8$), and $B$ a brick (we have $7$ bricks left after removing {$RR, R$}). So we have:
$ ^8C_2 * 7!(2!2!2!) = \frac {7*7!}{2}$
perms in total, since we choose $2$ of the $8$ spaces to distribute {$RR, R$} and we have $7!(2!2!2!)$ perms for the remaining $7$ bricks.
However, the answer provided is $7*7!$. Can anyone spot my error ?