In how many ways can the letters of the word ASSASSINATION be arranged so that no two vowels are together?

Question

$$\fbox{X}A\fbox{X}A\fbox{X}A\fbox{X}I\fbox{X}I\fbox{X}O\fbox{X}$$

Now, there are $\frac{6!}{3!2!}$ ways to arrange the vowels as I have placed them in the above schematic. Also, consonants will be fixed iin the places marked $X$. Again, there are $\frac{7!}{4!2!}$ ways to arrange the consonants. So, there are

$$\frac{6!}{3!2!}\cdot\frac{7!}{4!2!}$$

ways to arrange the letters of the word ASSASSINATION so that no two vowels are together. Am I correct?

Answer 1

You can arrange the consonants $SSSSNTN$ in $\frac{7!}{4!2!}$ ways.

Independently of this, you can insert the vowels $AAIAIO$ in the eight slots before, between and after each vowel. However that means two slots are unfilled - let’s denote these as $XX$.

Now you can arrange the slot-fillers $AAIAIOXX$ in $\frac{8!}{3!2!2!}$ ways.

Multiply these together and you have your total number of arrangements where no two vowels are together.

Answer 2

I expected you to try to correct your attempt; but as a solution has been posted, I will show another method which can be applied to any consonant-vowel permutation problem with a non-adjacent clause.

• Consider the $7$ consonants as blue balls, the $6$ vowels as red balls, and place the consonants one by one followed by placing vowels one by one.

• Each consonant placed creates an extra placement possibility for the next ball, but each vowel placed "eats up" one place for the next one due to the non-adjacent clause, thus $$\dfrac{(1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7)(8\cdot7\cdot6\cdot5\cdot4\cdot3)}{4!2!3!2!} =176,400$$

the denominator, of course, correcting for multiple letters of a kind

Answer 3

Since no one has (yet) highjacked the analysis in lulu's comment:

Normally, I would consider Inclusion-Exclusion for this type of problem. Here however, the Stars and Bars approach is simply much easier.

Any satisfying string of characters will be formatted like:

_V_V_V_V_V_V_

where the V's represent the 6 vowels.

Let $~A~$ denote the number of ways that the $~6~$ vowels can be permuted.

Let $~B~$ denote the number of ways that the $~7~$ consonants can be permuted.

Let $~C~$ denote the number of different ways of satisfactorily spacing the vowels.

Then, the desired computation is

$$A \times B \times C.$$

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Note
Suppose that $~n_1, ~n_2, ~n_3, ~$ are each positive integers, and that you have a Multi-Set that contains:

• An element $~r~$ with multiplicity $~n_1.~$
• An element $~s~$ with multiplicity $~n_2.~$
• $~n_3~$ other elements, each with a multiplicity of $~1.$

Then, the number of ways of permuting the elements in this multiset is

$$\frac{(n_1 + n_2 + n_3)!}{(n_1!) \times (n_2!)}.$$

Therefore,

$$A = \frac{6!}{3! \times 2!}, ~B = \frac{7!}{4! \times 2!}.$$

So, the entire problem has been reduced to enumerating $C.$

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Consider the enumeration of the number of satisfying solutions to the following problem.

• $x_1 + x_2 + \cdots + x_7 = 7.$

• $x_1, x_7 \in \Bbb{Z_{\geq 0}}.$

• $x_2, x_3, \cdots, x_6 \in \Bbb{Z_{\geq 1}}.$

The above constraints perfectly capture the requirement that none of the $~6~$ vowels are next to each other. That is, $~x_1~$ and $~x_7~$ are each allowed to equal $~0,~$ which corresponds to the fact that both the first character and the last character of the sequence of characters are permitted to be a vowel.

Then, the requirement that each of $~x_2, \cdots, x_6~$ be $~\geq 1~$ captures the fact that none of the vowels may be next to each other.

For Stars and Bars theory, see this article and this article.

Per Stars and Bars theory, the number of solutions to:

• $x_1 + x_2 + \cdots + x_k = n ~: ~n \in \Bbb{Z^+}$

• $x_1, x_2, \cdots, x_k \in \Bbb{Z_{\geq 0}}$

is $~\displaystyle \binom{n + [k-1]}{k-1}.$

So, the Stars and Bars problem that is being used to compute the value $C$ must be converted into the above format, where each of the variables has the constraint that $~x_i \in \Bbb{Z_{\geq 0}}.$

In the Stars and Bars computation of C, the variables $~x_1,~$ and $~x_7~$ are already okay.

To adjust the other variables, $~x_2, \cdots, x_6,~$ use the following change of variable:

For $~i \in \{2,3,\cdots,6\},~$ let $~y_i = x_i - 1.$

Therefore, for $~i \in \{2,3,\cdots,6\},~$, you will have that :

• $y_i \in \Bbb{Z_{\geq 0}} \iff x_i \in \Bbb{Z_{\geq 1}}.$

• $x_1 + x_7 + (y_2 + \cdots + y_6) = 7 - 5 = 2 \iff $
  $x_1 + x_7 + (x_2 + \cdots + x_6) = 7.$

Therefore, the use of Stars and Bars theory to computate C has been converted to enumerating the number of solutions to

• $x_1 + x_7 + (y_2 + \cdots + y_6) = 2$

• $x_1, x_7 \in \Bbb{Z_{\geq 0}}.$

• $y_2, \cdots, y_6 \in \Bbb{Z_{\geq 0}}.$

Therefore, per Stars and Bars theory,
$C = \displaystyle \binom{2 + 6}{6} = \binom{8}{6} = 28.$

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$\underline{\text{Final Computation}}$

$$A \times B \times C = \frac{6!}{3! \times 2!}, \times \frac{7!}{4! \times 2!} \times 28 = 176400.$$