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How many arrangements are there of TINKERER with two, but not three consecutive vowels?


  • Possible number of pairs: 3 (EE, EI, IE)
  • Number of arrangements for end cases (when the pair is at end or beginning): $2\cdot5\cdot\frac{5!}{2!} = 600$, where the $2!$ is to adjust for the over-counting of a repeated "R".
  • For middle cases, there are $5\cdot4\cdot\frac{5!}{2!} = 1200$ arrangements. This is because the surrounding letters of, say, "EE", cannot be "I". So there are there are $5$ possible choices (instead of $6$) of letters before EE and $4$ after (instead of $5$).

Summing the middle and end cases, then multiplying by the number of pairs, we have $5400$. $\square$

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up vote 4 down vote accepted

I would like to propose a different approach to this question - Consider if you arrange like this $$ T \space N\space K\space R\space R$$ There are, indeed $\frac {5!}{2!}$ ways to arrange, with $2!$ to account for double-counting $R$.

You mentioned that there are $6$ slots, which is correct. You can choose any $2$ of the $6$ slots to insert your 'blocks' of vowels. Total number of ways here will be $ 6 \choose 2 $ ways.

There are $3$ different ways to make the vowels.

Finally, when you slot them in, there are $2$ ways. Eg. TNK$EE$R$I$ is different from TNK$I$R$EE$ - selecting the same slot, but changing the way the blocks of vowels are inserted.

Final answer is, in order of paragraphs,
$$ {\frac {5!}{2!}} {\times} {6 \choose 2} \times 3 \times 2 = 5400 $$

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Good cases analysis: your solution and answer are correct. A small modification may be a little easier. Start by pretending the letters are all different.

By a cases analysis like yours, there are $30$ ways to choose the slots the vowels will go into. There are then $3!$ ways to fill these slots with vowels, and for each way of doing that, there are $5!$ ways of filling the remaining slots with consonants.

Finally, divide by $2!2!$ to take account of the fact there are $2$ e's and $2$ r's.

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How do you obtain $30$? – AlanH Feb 3 '13 at 3:22
One can do it using your reasoning. Left end for the double. That gives $5$ places for single, right end same, so $10$. Not an end for double, $5$ choices for left end of double. For each of these, $4$ choices for single, total $20$. So $10+20$. There are better ways. – André Nicolas Feb 3 '13 at 3:29

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