It is given that the number $43361$ can be written as a product of $two$ distinct prime numbers $p_1$ and $p_2$. Further,assume that there are $42900$ numbers which are less than $43361$ and are coprime to it. Then,the value of $p_1+p_2$ is

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:

The coefficient of $x^7$ in $(1-x+2x^3)^{10}$ is $........$.

If $(1-x^3)^{10} = \sum_{r=0}^{10} a_r x^r (1-x)^{30-2r}$, then $\frac{9a_9}{a_{10}}$ is equal to . . . . . . .

Let $d(n)$ denote the number of divisors of $n$ including $1$ and itself. Then, $d(225)$, $d(1125)$, and $d(640)$ are

The number of divisors of $7!$ is