Divisors of $9!$ which are perfect square, cube or power $4$ [1] Total no. of positive divisors of $9!$ which are perfect square.
[2] Total no. of positive divisors of $9!$ which are perfect cube.
[3] Total no. of positive divisors of $9!$ which are perfect power 4.
$\underline{\bf{My\; Try}}$:: prime factors of $9! = 2^7 \times 3^4 \times 5 \times 7$
[1] Total no. of Divisors which are perfect square $9! = (2^3)^2\cdot (3^2)^2$ is $ = (2+1)\cdot (3+1) = 12$
[2] Total no. of Divisors which are perfect cube $9! = (2^2)^3\cdot (3^3)^1$ is $ =(3+1)\cdot (1+1) = 8$
[3] Total no. of Divisors which are perfect cube $9! = (2^4)^1\cdot (3^4)^1$ is $ =(1+1)\cdot (1+1) = 4$
Now My Question is that is it Right or not. If not please explain me
Thanks

We use your prime factorization. Note that the perfect squares will have to be of shape $2^{2m} 3^{2n}$, where $m$ and $n$ are non-negative integers, and $2m\le 7$, $2n\le 4$. 
For $m$ we can choose any of $0$, $1$, $2$, or $3$ ($4$ choices). For each of these choices, we can let $n=0$, $1$, or $2$. Thus the number of square divisors is $(4)(3)$, your answer.
For perfect cubes, our divisor will have shape $2^{3m}3^{3n}$. There are $3$ choices for $m$, and $2$ choices for $n$, giving a total of $(3)(2)$.
Fourth power is similar. We get $(2)(2)$. 
