<s>[INST] 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
 [/INST] 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 $(2)(2)$. 
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