# Number of arrangements such that no vowels are together

Find number of arrangements of the word $EAMCOT$ such that no two vowels are adjacent to each other.

So,my book gives the answer $144$.

I tried to do it this way-$$\text{_ E _ A _ O _ }$$However I cannot fill the blanks in $4P3$ $\times3!=144$ ways with the remaining letters as there might be words like $\text{CEAMOT}$ in which two vowels are adjacent.

So,I tried to separate two cases like this-$$\text{_ E _ A _ O and E _ A _ O _}$$ which can be arranged in total $(3!\times3!)\times2=72$ ways.

Am I wrong?

• Your book has the best approach, otherwise. You missed out on 2 cases. ( as pointed out by Goh) that should give you (6*6)*4 – The Dead Legend Oct 18 '17 at 5:19
• @TheDeadLegend I think the first method is incorrect for the reason I cited... – tatan Oct 18 '17 at 5:22
• please see that E,A and O are always fixed on the three available spots. we are just moving around C,T,M using 4P3 on the three blanks. In this case vowels coming together is Simply out of Question! – The Dead Legend Oct 18 '17 at 5:25
• @TheDeadLegend I understand.But what about $\text{CE _ AMOT}$. Here $E,A$ come together.Don't they? – tatan Oct 18 '17 at 5:27
• @TheDeadLegend A better method would be to find all possible cases and then subtract the complement. – tatan Oct 18 '17 at 5:34

You left out two cases which are of the form of

$$A**E*O$$

$$A*E**O$$

Hence $$(3!\times 3!)\times 4$$

Here is an approach for separating the vowels which eliminates the need to consider cases.

We will arrange three blue balls and three green balls in a row, with the green balls separated. Line up the three blue balls. This creates four spaces. $$\square \color{blue}{b} \square \color{blue}{b} \square \color{blue}{b}\square$$ To separate the three green balls, choose three of these four spaces in which to place a green ball. For instance, if we choose the first, second, and fourth spaces, we obtain the arrangement $$\color{green}{g}\color{blue}{b}\color{green}{g}\color{blue}{b b}\color{green}{g}$$ Arrange the three consonants in the three positions occupied by the blue balls and the three vowels in the three positions occupied by the green balls. Hence, there are $$\binom{4}{3}3!3! = 144$$ arrangements of the six letters in which the three vowels are separated.

• Just a suggestion on presentation: it's hard for me to see the difference between the blue and green dots in your second display (and a significant proportion of the population is color-blind, for that matter). It might be better to choose some other method of displaying the difference between the dots. Good solution, thanks. – awkward Oct 18 '17 at 14:01
• @awkward I replaced the bullets with the letters $b$ and $g$. Does that make it clearer? – N. F. Taussig Oct 18 '17 at 14:10
• Yes, that's a great improvement in legibility. Thank you. – awkward Oct 18 '17 at 14:22