Can you please explain how you got that?
I calculated and found answer to be $24/105$ as follow:
$Total\space possible \space permutation = 9! / (2!\space 3! \space 4!)$
$Total\space possible\space permuations\space that\space we\space want = 2!\space 3!\space 4!$
So $final\space probabilty = (2!\space 3!\space 4!) / (9! / (2!\space 3! \space 4!)) = 24/105$
Can you suggest where I got wrong.