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Question

How many different ways can I arrange the letters in BOOKKEEPER if vowels must appear in alphabetical order?

How many different ways can I arrange the vowels? That would be 1 way. No I glue the term $eeeoo$ together and treat it as a letter of its own. That would leave me with $6$ letters left. Different ways to arrange them? $${6! \over 2!}$$ since there are $2 \;k's$. Did I do it right? Thank you in advance for your time.

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    $\begingroup$ Presumably you meant to write $6!/2!$, not $6/2!$. That answer is correct if you interpret the instructions as keeping the vowels together in alphabetical order. But if you allow arrangements such as KEBEPEKORO, you get a larger number. $\endgroup$ – Barry Cipra Sep 23 '15 at 21:51
  • $\begingroup$ I completely didn't account for that, how do I go about it then... $\endgroup$ – ponderingdev Sep 23 '15 at 22:02
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Method 1: We use symmetry.

For the moment, let's consider the number of distinguishable ways we can permute the ten letters in BOOKKEEPER. We can place the three E's in three of the ten locations in $\binom{10}{3}$ ways. We can place the two O's in two of the remaining seven places in $\binom{7}{2}$ ways. We can place the two K's in two of the remaining five places in $\binom{5}{2}$ ways. There are then $3!$ ways of arranging the B, P, and R in the three remaining places. Hence, the number of distinguishable arrangements of BOOKKEEPER is $$\binom{10}{3}\binom{7}{2}\binom{5}{2} \cdot 3! = \frac{10!}{7!3!} \cdot \frac{7!}{5!2!} \cdot \frac{5!}{3!2!} \cdot 3! = \frac{10!}{3!2!2!}$$

Now, let's restrict our attention to arrangements of the five vowels in BOOKKEEPER. Since there are three E's and two O's, a given permutation of EEEOO is determined by in which three of the five positions the E's are placed. There are $\binom{5}{3} = 10$ ways to do this, of which just one is in alphabetical order.

Hence, the number of permutations of the letters of BOOKKEEPER in which the vowels appear in alphabetical order is $$\frac{1}{10} \cdot \frac{10!}{3!2!2!} = \frac{9!}{3!2!2!}$$

Method 2: We place the consonants first.

There are $\binom{10}{2}$ ways of choosing the positions of the two K's, eight ways to place the B, seven ways to place the P, and six ways to place the R. Once the consonants have been placed, there is only way to fill the five remaining positions with the vowels in alphabetical order. Hence, the number of distinguishable arrangements of the letters of BOOKKEEPER in which the vowels appear in alphabetical order is $$\binom{10}{2} \cdot 8 \cdot 7 \cdot 6$$

Method 3: We place the vowels first.

There are five vowels in BOOKKEEPER, which has ten letters. We can select positions for the five vowels in $\binom{10}{5}$ ways. There is only one way to arrange the vowels in those positions in alphabetical order. There are $\binom{5}{2}$ ways to place the K's in two of the remaining five positions. There are $3!$ ways to arrange the B, P, and R in the remaining three positions. Hence, the number of distinguishable arrangements of BOOKKEEPER in which the vowels appear in alphabetical order is $$\binom{10}{5}\binom{5}{2} \cdot 3!$$

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    $\begingroup$ (+1) There are some differences in our answers, but if you think they are too close, I will remove mine since yours was here first. $\endgroup$ – robjohn Sep 23 '15 at 22:46
  • $\begingroup$ @robjohn (+1) for your answer since you provided alternate ways of looking at both parts of my answer. $\endgroup$ – N. F. Taussig Sep 23 '15 at 22:49
  • $\begingroup$ I don't understand why you divide by 10 though..are you treating it like a probability problem? sorry if i just don't see it $\endgroup$ – ponderingdev Sep 23 '15 at 23:03
  • $\begingroup$ For a given permutation of the consonants, there are ten ways of arranging the vowels, of which just one is in alphabetical order. Thus, one tenth of the total number of permutations is in alphabetical order. Yes, you could view this argument as the probability that the vowels are in alphabetical order is $1/10$. $\endgroup$ – N. F. Taussig Sep 23 '15 at 23:09
  • $\begingroup$ the alternate method makes more sense to me in an instant! $\endgroup$ – ponderingdev Sep 23 '15 at 23:44
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One Approach

Without the alphabetical vowel constraint, there are $$ \frac{10!}{\underbrace{1!}_{\text{B}}\underbrace{2!}_{\text{O}}\underbrace{2!}_{\text{K}}\underbrace{3!}_{\text{E}}\underbrace{1!}_{\text{P}}\underbrace{1!}_{\text{R}}} $$ Arrangements of the letters. For each arrangement with the vowels in alphabetical order, there are $\binom{5}{2}$ unconstrained arrangements of the vowels (just count the arrangements of "EEEOO", of which there is only one that is in alphabetical order).

Thus, there are $$ \frac1{\binom{5}{2}}\frac{10!}{1!2!2!3!1!1!}=15120 $$ arrangements of the letters where the vowels are in alphabetical order.


Another Approach

Consider the vowels as one letter. This works since once we have selected where the vowels go, we know the order of the individual vowels within the group of vowels. Thus, there are $$ \frac{10!}{\underbrace{1!}_{\text{B}}\underbrace{5!}_{\begin{array}{c}\text{v}\\[-4pt]\text{o}\\[-4pt]\text{w}\\[-4pt]\text{e}\\[-2pt]\text{l}\\[-4pt]\text{s}\end{array}}\underbrace{2!}_{\text{K}}\underbrace{1!}_{\text{P}}\underbrace{1!}_{\text{R}}}=15120 $$ arrangements of the letters where the vowels are in alphabetical order.

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Imagine the letters as Scrabble tiles. Lay down the tiles EEEOO in the required alphabetical order. Now take the first consonant and place it in any of the $6$ positions before, between, or after what's there. Then take the first K and place it in any of the now $7$ positions before, between, or after what's on the table. Repeating this with the second K in any of the $8$ possible positions, the P in any of the $9$, and the R in any of the $10$ positions, we see there are $6\cdot7\cdot8\cdot9\cdot10$ ways of creating the arrangements. Finally, dividing by $2$ (or $2!$, if you like) to account for the two K's, we have the answer

$${6\cdot7\cdot8\cdot9\cdot10\over2}=15{,}120$$

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Another method is to count the number of ways of rearranging the symbols $vvvvvccccc$ where $v$ means vowel and $c$ means consonant. That's $10 \choose 5$. Then for each such arrangement count the number of ways of rearranging the vowels. That's $1$. Then count the number of weighs to rearrange the consonants. That's $5! \over 1!2!1!1!$. Finally multiply: ${10 \choose 5} \times 1 \times {5! \over 1!2!1!1!} = 15120$

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