JEE Math Practice Question
If $${({}^{30}{C_1})^2} + 2{({}^{30}{C_2})^2} + 3{({}^{30}{C_3})^2}\, + \,...\, + \,30{({}^{30}{C_{30}})^2} = {{\alpha 60!} \over {{{(30!)}^2}}}$$ then $$\alpha$$ is equal to :
- A.30
- B.10
- C.15
- D.60
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