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JEE MATHEMATICS MCQ: COMBINATIONS

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Application +4 | -1

If

\left({ }^{30} C_{1}\right)^{2}+2\left({ }^{30} C_{2}\right)^{2}+3\left({ }^{30} C_{3}\right)^{2}+\ldots+30\left({ }^{30} C_{30}\right)^{2}

=\frac{\alpha 60 !}{(30 !)^{2}}

then

\alpha

is equal to :

View answer and explanation

Correct option: C

15

Option C (15) is correct because the given expression can be simplified using combinatorial identities to find the value of alpha.

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