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Q: How many ways to arrange BOOKKEEPER where two E’s appear consecutively but not three.

Here What I've got : a) We can consider the two consecutive E’s as one block say X. Hence, we get a new string: XBOOKKPER of length 9. Therefore, the number of possible rearrangements for that word is obviously:

$$ \frac{9!}{2! \cdot 2!}$$

Then I need to remove the instances when there are three consecutive E's. There are two different ways of doing this which give me different answers, and I would like to understand which is correct. Way 1: To find "EEE", i can look at adding an e to my block X, and create a superblock Y. So Y = (e, X) or (X,e), two ways so I multiply by two how many arrangements of YBOOKPR so we get:

$$ 2\cdot\frac{8!}{2!\cdot 2!} $$

Way 2: Treat Y just being "EEE" and so we subtract only:

$$ \frac{8!}{2!\cdot 2!} $$

Tl;dr is the answer :

$$ \frac{9!-2 \cdot 8!}{2!\cdot 2!} \quad \text{or} \quad \frac{9!-8!}{2!\cdot 2!} $$

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  • $\begingroup$ Since $9!=9\cdot 8!$, the first (correct) way can be written conveniently as $\frac{(9-2)\cdot 8!}{2!\cdot 2!}=\frac{7\cdot 8!}{2!^2}$. $\endgroup$ Commented Nov 19 at 1:57

3 Answers 3

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When in doubt, test it out.

You can easily count the ways to so arrange "$\rm ZEEE$" and compare that to the results of your methods.

  • There are but $2$ ways to arrange it so that exactly two $\rm E$ are contiguous.

  • Yet there are $3!$ ways to arrange $\rm ZXE$, and $2!$ ways to arrange $\rm ZY$.

  • Clearly $2 = 6-2\cdot 2$, so counting $\rm Y$ as having $2$ arrangements ($\rm EX$ or $\rm XE$) is the correct method.

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The first way is correct; by grouping the two $E$s into a single block, a spelling containing $(EE)E$ would be counted separately from a spelling containing $E(EE)$.

In other words, you are actually overcounting the number of spellings which contain at least two consecutive $E$s in the first step (for precisely the reason that $(EE)E$ and $E(EE)$ are counted as different), but then you are correctly removing the redundant spellings by totaling all of the ways in which either $E(EE)$ or $(EE)E$ can appear, and then subtracting that from the total.

The second method does not account for this double-counting in the first step as it only counts the number of ways that $EEE$ can appear (and it does NOT distinguish between $E(EE)$ and $(EE)E$).

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You can use $\texttt{successive placement}$ to automatically avoid miscounting.

Make a block of two $E's$ as $\boxed{EE}$, successively place "tokens" with the third $E$ last (shown in red) avoiding block adjacency, and account as usual for duplicates.

The answer comes as $\dfrac{2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot\color{red}7}{2!2!} = \dfrac{8!\times 7}{2!2!}$

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