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Question

Words are formed using all letters of the word 'JEEADVANCED'.
Let a denotes the number of words in which all the vowels are together.
Let b denotes the number of words in which vowels as well as consonants are separated.
Let c denotes the number of words which begin and end with vowels.

A
c>a>b
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B
c>b>a
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C
a+b+c=460×6!
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D
a+c=91b
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Solution

The correct option is D a+c=91b
We have 11 letters
A,A,E,E,E,J,D,D,V,N,C,

If all vowels are together, then we will put them in a box and consider them as a single object. Now, we have 6 consonants and 1 object (all vovels). So, arrangements of the 7 objects =7!2! (D occurs twice)
Internal arrangement of the vowels
=5!2!3!a=7!2!×5!2!3!

For b, First we will arrange 6 consonants. To make them seperated we need to five 5 spaces among them by arranging vowels.
Arrangement of consonants
=6!2!
Arrangement of vowels
=5!2!3!b=6!2!×5!2!3!

For c, there are three cases
When the end letters are A,A
Number of words formed
=9!2!3!
When the end letters are E,E
Number of words formed
=9!2!2!
When the end letters are E,A
Number of words formed
=9!2!2!×2!
c=9!2!3!+9!2!2!+9!2!

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