MCQ
Let $X =\left({ }^{10} C _1\right)^2+2\left({ }^{10} C _2\right)^2+3\left({ }^{10} C _3\right)^2+\ldots \ldots . .+10\left({ }^{10} C _{10}\right)^2$ where ${ }^{10} C _{ r }, r \in\{1,2, \ldots ., 10\}$ denote binomial coefficients. Then, the value of $\frac{1}{1430} X$ is. . . . . . .
- A$430$
- B$435$
- C$540$
- ✓$646$