First, we'll place the vowels
& we can place the $5$ vowels in $5!$ ways
Now, $X$ can't be seated or placed in extreme position.
So,
Now, between any $2$ vowels, there can be at least $3$ X's
So, we're done with $3 \times 4 = 12 $ X's
& $12$ X's can be placed in $4$ ways (as all X's are identicals)
Now, we're left with $(15-12) = 3$ X's to place
Now, these $3$ X's can be placed in many ways-
- When all the $3$ X's are placed between the same $2$ vowels
∴ Total ways = $4$
- Place $2$ X's between the same pair of vowels, & place the remaining X in other pair of vowels
∴ Total ways = $4 \times 3$ $\qquad [∵\text{Once we fill 2 X's, then we'll have only 3 ways to fill 1 X}]$
- Place X's in different pair of vowels
∴ Total ways = $4$
Or, we can compute it in this way, ------
We have $4$ possibilities left & we have to place $3$ X's.
So, $^4C_3 = \dfrac{4!}{3! \times (4-1)!} = \dfrac{4!}{3!} = 4$
$\color{green}{\text{The total no. of ways} = 5! \times \left(4+ 4 \times 3 + 4 \right)}$
$\qquad \qquad = 120 \times \left(4+12+4 \right)$
$\qquad \qquad = 120 \times 20$
$\qquad \qquad = \color{gold}{2400 \hspace{0.1cm} ways}$
Correct Answer: $C$