Find the coefficient of $x^{64}$ in the expansion of $P(x) = (x - 1)^2(x - 2)^3(x - 3)^4 \dots (x - 10)^{11}$.

If $(1+x)^{15}=a_0+a_1 x+\ldots+a_{15} x^{15}$,then $\sum_{r=1}^{15} r \frac{a_r}{a_{r-1}}$ is equal to

Sum of the coefficients of $x^4$ and $x^6$ in the expansion of $(1+x-x^2)^6$ is

In the binomial $(2^{1/3} + 3^{-1/3})^n$,if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is $1/6$,then $n =$

If $n$ is the degree of the polynomial,$\left[ {\frac{1}{{\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8 + \left[ {\frac{1}{{\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8$ and $m$ is the coefficient of $x^{12}$ in it,then the ordered pair $(n, m)$ is equal to

The value of $x$ in the expression $(x + x^{\log_{10} x})^5$,if the third term in the expansion is $1,000,000$.