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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 ?

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    Provided by the textbooks from which I happen to be randomly selecting questions. Why?2011-03-01

1 Answers 1

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It is not enough to just consider the $ ^8C_2$ locations where the elements $RR$ and $R$ will go. The arrangement is also important. So, if you were to exchange $RR$ with $R$ in a given permutation, you would get a different permutation. So you have to multiply your answer by $2$ to get $7*7!$.

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    Thanks. Of course. In this case we need 8P2 for the first term. It's incredibly easy to miss these things. I'd forgotten how subtle combinatorial problems can be, even elementary ones.2011-03-01