Among the positive divisors of the number $12600$,if $n_1$ is the number of divisors which are multiples of $3$ and $n_2$ is the number of divisors which are multiples of $14$,then $n_1 + n_2 =$

Consider the following statements:
$I$: The number of non-trivial even divisors of the number $N = 2^{\alpha_1} 3^{\alpha_2} 4^{\alpha_3} 5^{\alpha_4} 6^{\alpha_5}$ is $(\alpha_1+2\alpha_3+\alpha_5)(\alpha_2+\alpha_5+1)(\alpha_4+1)-1$.
$II$: The number of non-trivial odd divisors of the number $N = 2^{\alpha_1} 3^{\alpha_2} 4^{\alpha_3} 5^{\alpha_4} 6^{\alpha_5}$ is $\alpha_2+\alpha_4+\alpha_5+\alpha_2\alpha_4+\alpha_4\alpha_5$. Then:

Three players play a total of $9$ games. In each game,one person wins and the other two lose; the winner gets $2$ points and the losers get $-1$ each. The number of ways in which they can play all the $9$ games and finish each with a zero score is

$A$ natural number has prime factorization given by $n = 2^{x} 3^{y} 5^{z}$,where $y$ and $z$ are such that $y+z=5$ and $y^{-1}+z^{-1}=\frac{5}{6}$,with $y > z$. Then the number of odd divisors of $n$,including $1$,is ..... .

What is the highest power of $3$ in $100!$?

The coefficient of $x^{10}$ in the expansion of $(1+x^2-x^3)^8$ is