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JEE MATHEMATICS MCQ: PERMUTATIONS

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\frac{(2 n) !}{2^{2 n}(n !)^2} \leq

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Correct option: A

\frac{1}{\sqrt{3 n+1}}

Option A (

\frac{{1}}{{\sqrt{{3n+1}} }}

) is correct because substituting n=1 into the expression results in equality, and for n>1 the inequality holds.

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