0
$\begingroup$

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 ?

  • 0
    "The answer provided" --- *by whom?*2011-03-01
  • 0
    You forgot to consider what happens if R goes first and RR goes second, vs. what happens if RR goes first and R goes second; you treated them as indistinguishable, they aren't.2011-03-01
  • 0
    Provided by the textbooks from which I happen to be randomly selecting questions. Why?2011-03-01

1 Answers 1