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. . . . . . .

  • [IIT 2018]
  • A

    $430$

  • B

    $435$

  • C

    $540$

  • D

    $646$

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