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. $c>a>b$
• B. $c>b>a$
• C. $a+b+c=460\times 6!$
• D. $a+c=91b$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $c>a>b$
C $a+b+c=460\times 6!$
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 $=\frac{7!}{2!}$ (D occurs twice)
Internal arrangement of the vowels
$=\frac{5!}{2!3!}\Rightarrow a=\frac{7!}{2!}\times \frac{5!}{2!3!}$

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
$=\frac{6!}{2!}$
Arrangement of vowels
$=\frac{5!}{2!3!}b=\frac{6!}{2!}\times \frac{5!}{2!3!}$

For $c$, there are three cases
When the end letters are A,A
Number of words formed
$=\frac{9!}{2!3!}$
When the end letters are E,E
Number of words formed
$=\frac{9!}{2!2!}$
When the end letters are E,A
Number of words formed
$=\frac{9!}{2!2!}\times 2!$
$\therefore c=\frac{9!}{2!3!}+\frac{9!}{2!2!}+\frac{9!}{2!}$