All the letters of the word "eamcot" are arranged in different possible ways.
Find the number of arrangements where no two vowels come together.
The approach I used –
Assume you have 6 slots.
You can fill the first slot in two broader ways – vowel or consonant.
According to this choice all the next places have already been decided because you can only go in an alternating manner due to the condition.
Now for filling all the consonants you can do that in 3! ways and vowels in 3! ways as well.
So the final answer is 3!*3!*2 = 72.
(The two is accounting for the 2nd step)
But all over the internet I am getting the answer as 144.
They do the following –
You know that any such permutation can be written as cvcvcvc (one of the "c"s is empty).
Now filling all "v"s is 3! and filling all "c"s is 4! (I think the problem here is that you can only have the empty space either on the start or end only and not anywhere else and hence 4! is incorrect) and they get the answer as 144.
Which of the two approaches is correct?
Best Answer
The problem states that no two vowels are adjacent. It does not say that no two consonants are adjacent. Therefore, your answer is wrong, as is the justification for the correct answer that you found on the internet.
Arrange the three distinct consonants in $3!$ ways. That creates four spaces in which we can place the three vowels, two between successive consonants and two at the ends of the row. $$\square C_1 \square C_2 \square C_3 \square$$ In the diagram, $C_i$ represents the position of the $i$th consonant and the squares represent the spaces where a vowel can be placed. To separate the vowels, we must choose three of these four spaces in which to place a vowel, which can be done in $\binom{4}{3}$ ways. The three distinct vowels can be arranged in the selected spaces in $3!$ ways. Hence, the number of arrangements of the letters of the word EAMCOT in which no two vowels are adjacent is $$3!\binom{4}{3}3! = 144$$