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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:

8C2 * 7!(2!2!2!) = 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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"The answer provided" --- by whom? –  Rasmus Mar 1 '11 at 15:38
    
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. –  Arturo Magidin Mar 1 '11 at 16:00
    
Provided by the textbooks from which I happen to be randomly selecting questions. Why? –  user7597 Mar 1 '11 at 16:04
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1 Answer

It is not enough to just consider the 8C2 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. –  user7597 Mar 1 '11 at 16:00
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