# Combinations word “GRACEFUL” question

The total letter arrangements for the word GRACEFUL is 8!.

The question is: What fraction of all arrangements of GRACEFUL have no pair of consecutive vowels?

Now the three vowels in this word are A, E, and U. They can appear consecutively in 3P3 ways so this can be handled in 8! / 3P3 ways. But it is also possible that any two can appear consecutively.

So would the answer be: 8! / (3P3) * 2! * 3?

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Please use the Inclusion-Exclusion Principle. –  hkju Apr 19 '12 at 20:02

Since there cannot be consecutive vowels, remove two consonants (we'll add them back in shortly). We can look at the number of ways to arrange $3$ vowels and $3$ consonants, now without the restriction on consecutive vowels. There are $6\choose3$ ways to do this. For any arrangement, add one consonant between the first and second vowel, and another between the second and third.

Now assign vowels and consonants: there are $(3!)(5!)$ ways to do this. The fraction you are looking for is therefore: $$\frac{{6\choose3}(3!)(5!)}{8!}$$

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Variant, inspired by Sasha's final answer: after finding $6\choose3$ as described above, now notice that there are $8\choose3$ ways in total to distribute $3$ vowels in an $8$-letter word. This gives $\frac{\binom{6}{3}}{\binom{8}{3}}$. –  Théophile Apr 20 '12 at 14:25

There are 3! ways to arrange vowels. Let there be $k_1$ consonants before the first vowel, $k_2$ between the first and the second, $k_3$ between the second and the third, and $k_4$ after the third vowel. Clearly: $$k_1+k_2+k_3+k_4 = 8 - 3 =5, \qquad k_1 \geqslant 0, k_4 \geqslant 0, k_2 \geqslant 1, k_3 \geqslant 1$$ Thus the number of arrangements: $$3! \cdot 5! \sum_{k_1=0}^3 \sum_{k_2=1}^4 \sum_{k_3=1}^4 \sum_{k_4=0}^3 \delta_{k_1+k_2+k_3+k_4,5} = 3! \cdot 5! \cdot 20$$ The sum can be evaluated as follows: $$\sum_{k_1=0}^3 \sum_{k_2=1}^4 \sum_{k_3=1}^4 \sum_{k_4=0}^3 \delta_{k_1+k_2+k_3+k_4,5} = \sum_{k_1=0}^3 \sum_{k_2=0}^3 \sum_{k_3=0}^3 \sum_{k_4=0}^3 \delta_{k_1+k_2+k_3+k_4,3} = [z]^3 \left( \frac{1-z^4}{1-z}\right)^4 = [z]^3 \left( \frac{1}{1-z}\right)^4 = \binom{3+4-1}{3} = \binom{6}{3}$$

The fraction is thus $$\frac{\binom{6}{3}}{\binom{8}{3}} = \frac{5}{14}$$

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You are right that the total number of arrangements is $8!$.

I don't understand your "$8!/3P3$" to "handle" the case of all three vowels appearing consecutively: if all three vowels appear together, then there are six locations for the block of three vowels to be placed at. There are $3!$ ways of placing the vowels in the block, and $5!$ ways of placing the remaining letters in the remaining spots. That gives $6\times 3!\times 5!$ ways in which the three vowels may appear together. But this is neither equal to $8!/3P3$, nor is it equal to $8!-8!/3P3$. So I don't understand what you mean by "handling" the case, or how the case relates to $8!/3P3$.

In how many ways can at least two vowels appear together? Select the location of the block of two vowels, which can be done in $7$ different ways. Then select the vowels to appear in the block, in order: $3\times 2$. Then place the remaining six letters in the remaining six spots: $6!$. We get $7\times 6\times 6!$. But this overcounts: if the three vowels appear together, then we counted that particular arrangement twice: once when we selected the first two vowels as the block, once when we selected the last two vowels as the block.

So: the total number in which no two vowels appear together would be: $$(\text{total number of arrangements}) - (\text{arrangements with at least two vowels together}) + (\text{arrangements will all three together}).$$

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