No. of ways 9 prizes can be given to 5 students with a condition 
Number of ways 9 prizes can be given to 5 students if one particular student receives 4 prizes and the rest of the students can get any number of prizes is?

My Attempt:
I got the possibilities for distribution of prizes as,
$(4,0,1,1,3)$
$(4,1,1,1,2)$
$(4,0,0,3,2)$
Solving further includes permutations of the prizes,
$5![\frac{9!}{4!2!3!} + \frac{9!}{4!3!2!} + \frac{9!}{4!2!3!2!}]$
However, solving this gives me a wrong answer.
Any help would be appreciated.

Following your approach, the distribution of prizes (we assume that they are distinguishable) should be
$$({\bf 4},1,1,1,2),({\bf 4},0,1,1,3),({\bf 4},0,1,2,2),({\bf 4},0,0,1,4),({\bf 4},0,0,2,3),({\bf 4},0,0,0,5)$$
and including the permutations of the remaining $5$ prizes and of the other $4$ students we obtain
$$\binom{9}{4}\left(\frac{4!}{3!1!}\cdot \frac{5!}{1!1!1!2!}+
\frac{4!}{1!2!1!}\cdot \frac{5!}{0!1!1!3!}+\dots+\frac{4!}{3!1!}\cdot \frac{5!}{0!0!0!5!}\right)=129024.$$
The same result can be obtained in a simpler way as
$$\binom{9}{4}\cdot 4^5=129024$$
where $\binom{9}{4}$ is the number of ways to assign $4$ prizes to the particular student and $4^5$ is the number of ways to give each of the remaining $5$ prizes to one of the other $4$ students. 
