How many ways can the letters of MATHEMATICS be arranged if no two vowels are adjacent?

Question

I am trying to do this question from a high school textbook. The answer given is 1058400, but I can't get this answer.

Here's what I've tried so far: There's 11!/(2^3) arrangements without any restrictions

If the two As are adjacent, then there's 10!/(2^2) arrangements (you can swap the As, but since they are repeated there's no need to)

Then you can have AE, AI, EI adjacent. In each case this is (10!*2!)/(2^3), as the two vowels can be swapped around, but the As are repeated.

Subtracting these cases from the number of arrangements without any restrictions gives 1360800, which is not the answer.

Answer 1

• All possible arrangements with A,T,M twice: $\color{blue}{\frac{11!}{2!\cdot 2! \cdot 2!}}$
• The $\color{blue}{7}$ consonants with T,M twice: $\color{blue}{\frac{7!}{2!\cdot 2!}}$
• These consonants written as list give rise to $\color{blue}{8}$ possible places where to put at most one vowel. So, choose $\color{blue}{4}$ of these places: $\color{blue}{\binom{8}{4}}$
• Now, distribute the 4 vowels into each selection of 4 of these places and note that A appears twice: $\color{blue}{}$ $\color{blue}{\frac{4!}{2!}}$

All together:

Number of ways with no two vowels adjacent:

$$\color{blue}{\frac{7!}{2!\cdot 2!} \cdot \binom{8}{4} \cdot \frac{4!}{2!} = 1058400}$$

Number of ways with at least two vowels adjacent:

$$\color{blue}{\frac{11!}{2!\cdot 2! \cdot 2!} - \frac{7!}{2!\cdot 2!} \cdot \binom{8}{4} \cdot \frac{4!}{2!} = 3931200}$$

Answer 2

trancelocation did exact calculations, but he confused to give the final answer.

Of course the number of all possible arrangements without any restrictions (taking into account that A,T,M appear twice) is: $\color{blue}{\frac{11!}{2! \cdot 2! \cdot 2!}}$. But it's irrelevant.

So the answer is:

(stating trancelocation's answer)

• The number of arrangements for the 7 consonants with T,M twice is: $\color{blue}{\frac{7!}{2! \cdot 2!}}$
• These consonants give rise to 8 possible places where to put at most one vowel. So, choose 4 of these 8 places: $\color{blue}{\binom{8}{4}}$
• Now, for each selection of the 4 vowels, the number of arrangements of the vowels themselves with A twice is: $\color{blue}{\frac{4!}{2!}}$

All together: $$\color{blue}{\frac{7!}{2! \cdot 2!} \cdot \binom{8}{4} \cdot \frac{4!}{2!} = 1,058,400}$$