Question

How many ways are there to arrange the letters in word G A R D E N with the vowels in alphabetical order?

Answer 1

In this type of problem we have to find the total number of ways in which a word can form and then divide it by the ways in which vowels can arrange, find the number of vowels, and consonants in the word. Or In 720 words half of them have vowels in alphabetical order and the other half are not in alphabetical order. Accordingly, find the required value.

Complete Step-by-step Solution
A word G A R D E N is given in which there are $6$ letters are used (G A R D E N).
We have to arrange the letters that vowels can only arrange in alphabetical order.
In the given word we have two vowels (A, E).
So, we have to arrange the word in a manner that A always comes before E.
These two can be arranged in 2 ways i.e. either $\left( {A,E} \right)or\left( {E,A} \right)$
$\Rightarrow 2! = 2$
The total numbers of ways by which $6$ letters (G A R D E N) can arrange to form a word are $6!$ ways.
For the arrangement of 6 words including two vowels, it is equal to $6! = 720$
But we have to choose the first order for vowel that is $\left( {A,E} \right)$
In 720 words half of them have vowels in alphabetical order and the other half are not in alphabetical order.
So the required answer will be equals to:
$\dfrac{{6!}}{{2!}}or\dfrac{{720}}{2}$
Now, solve this to find out the number of ways.

$$= \dfrac{{6 \times 5 \times 4 \times 3 \times {{2 \times 1}}}}{{{{2 \times 1}}}} \\\ = 360 \\\$$

So, there are $360$ ways in which letters of the word “G A R D E N” can arrange with the vowels in alphabetical order.

Note:
Total ways come $6!$ because a letter cannot be repeated to form a word with meaning or without meaning and the vowels are given (A, E) only come simultaneously ( it means A always comes before E in the sequence).