Question

Find the number of ways in which letters of the word VALEDICTORY be arranged so that the vowels may never be separated.

Answer 1

Count the total number of vowels present in the word VALEDICTORY. Now make those vowels as a pack of obe and permute the rest of the letters by taking the vowels as a single entity.

Complete step by step answer:
So here in this word 4 vowels are present and those are A, E, I, O
Now let us assume that these vowels are basically a single entity
We can clearly see that total number of letters in the word VALEDICTORY without taking the vowels as one is 11.
Now after taking the the vowels as 1 we will be getting the number of letters as $(11 - 4) + 1 = 8$
As there were 11 letters and we subtracting 4 vowels so that's why $(11 - 4)$ and we were taking 1 letter to replace the vowels so $(11 - 4) + 1$ which eventually comes out to be 8
So there are a total of $8!$ ways to permute 8 letters in a word
Now we may also observe that as we were taking a Pack of vowels as one entity then the letters within that entity will also permute.
Clearly there are 4 letters within the vowel entity so it will permute in $4!$ ways.
Hence we will get the required permutation as $8!(4!)$

So, the correct answer is “ $8!(4!)$”.

Note: Do note that there are a total of 5 vowels in the alphabetical system that are as follows: A, E, I, O, U. In this type of questions always take something as one entity and then try to solve further.