Number of Derangements of the word BOTTLE I am wondering how to calculate the number of derangements for the word BOTTLE. I understand how to actually do the formula for derangements already. My issue is what do you do with repeated letters. Obviously, I will be over counting if I do the typical formula. Makes me think it is the number of derangements with the letter T in their original space, but I am not sure. Can anyone help, as I am wondering if I am supposed to use PIE to solve this. Thanks. 
 A: There are $\left\lfloor \frac{6!}{e}\right\rfloor = 265$ derangements of a 6-element set, but applying that formula to the letters of "BOTTLE" has two problems:


*

*Some of these "derangements" move the T in the fourth position to the third position, or vice versa, or both, so once we take that into account, they're no longer actual derangements.

*Of the valid derangements, each is counted twice: if you switch the two T's, that's a different permutation of a $6$-element set, but shouldn't be a different derangement of the letters of "BOTTLE".
We begin by fixing the first problem. There are three cases:


*

*The false derangement actually swaps the two T's. This case is in bijection (by swapping the two T's) with permutations that fix both T's and derange everything else, so there are $\left\lfloor \frac{4!}{e}\right\rfloor = 9$ of these.

*The false derangement moves the first T to the second T's place, and the second T to somewhere other than the first T's place. This case is in bijection (by swapping the two T's) with permutations that fix the second T and derange everything else, so there are $\left\lfloor \frac{5!}{e}\right\rfloor = 44$ of these.

*Same as the previous case, but with the second T going to the first T's place; also $44$ of these.


This leaves us with $265-44-44-9 = 168$ actual derangements.
The second problem is easy to fix; now we can divide by $2$ and get $84$ as our final answer.
(I also cheated and confirmed this by brute force in Mathematica.)
A: Choose two valid locations for the letters $T$, $\binom 42$ ways. The letters that started at those locations are "free", the other two are "unfree".
Choose a location for the first unfree letter from the the $3$ available to it. Two cases:  


*

*choose one of the two $T$ start positions, in which case the other unfree letter has only two choices, or  

*choose the other unfree letter start position, in which case that letter is free. 


Remaining $n$ free letters go as $n!$.
Over all we have $\binom 42(2\cdot2\cdot 2! + 1\cdot 3!) = 6\cdot (8+6) = 84$.

Steampunk Solutions Inc
A: There are four letters other than the $T$s. Thus there are $4 \cdot 3 = 12$ choices for what to put under the two $T$s. There are $\dfrac{4!}{2!} = 12$ ways to arrange the remaining 4 letters (of which 2 are identical) in the four remaining slots, but the two non $T$s are in danger of being put in their original positions. There are $\dfrac{3!}{2!} = 3$ ways of putting the first of them (or the second) into its original position, and $\dfrac{2!}{2!} = 1$ way of putting both in their original positions. Thus the answer is $12(12-3-3+1) = 84$.
