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.