Question

Number of permutations of word ‘AUROBIND’ in which vowels appear in an alphabetical order is:
A) \[P(8,4)\]
B) \[C(8,4)\]
C) \[4!.C(8,4)\]
D) \[C(8,5).5!\]

Answer 1

Hint:
For this question, first arrange the vowels in the alphabetical order and note it down. Then, arrange the rest of the alphabets, count the number of places (which is the number of options for arranging the vowels), then insert any vowel and count the number of places again and keep on doing the same procedure until the end of the list of vowels is reached. In this question, we will use Combination.

Formula Used:
In this question, the formula for the Combination will be used, which is:
\[{}^n{C_r} = \dfrac{{n!}}{{n!\left( {n - r} \right)!}}\]

Complete Step by Step Solution:
In the given question, the word given to us is AUROBIND, which has a total 8 alphabets of which 4 are consonants, 4 are vowels, and all are unique.
Now, let’s take just the vowels and arrange them in alphabetical order –
\[A{\rm{ }}I{\rm{ }}O{\rm{ }}U\]
Now, let’s take the consonants and arrange them like they occur and leave the spaces too. The spaces will be marked by \[ \otimes \]
So, we have
\[ \otimes R \otimes B \otimes N \otimes D \otimes \]
Now, to place the first vowel, there are 5 options; let’s take the vowel A and place it to the left of N,
\[ \otimes R \otimes B \otimes A \otimes N \otimes D \otimes \]
Now, for the next vowel, there are 6 options; let’s take I and place it to the left of D,
\[ \otimes R \otimes B \otimes A \otimes N \otimes I \otimes D \otimes \]
For the next there are 7 options; let’s take O and place it to the right of D
\[ \otimes R \otimes B \otimes A \otimes N \otimes I \otimes D \otimes O \otimes \]

Now finally, there are 8 options for U
So, the total number of options for the vowels is - \[5 \times 6 \times 7 \times 8\]

Now, if we directly want to calculate the result, we can do so by applying the formula of Combination, but, we will have to multiply it by \[4!\] as among themselves, the consonants can be arranged in \[4!\] ways:
Total number of ways = \[{}^8{C_4} \times 4! = 4!.C\left( {8,4} \right)\]

Hence, the correct option is C.

Note:
So, we can see that in questions like these, we can solve and derive the logic for the same by manually putting in what is asked in the question. It is very important that we always keep in mind that when we have a question where the order does not matter, we use Combination but when the order does matter, we use Permutation. It is the most basic thing in Permutation and Combination and it is upon this concept how all the other formulae and concepts were derived.