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If $n > 2, r > 1$ and the coefficients of $(3 r)$ th and $(r + 2)$ th terms in the expansion of ${(1 + x)}^{2 n}$ are equal, then

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$n = 2 r$

$n = 3 r$

n = 2r + 1

n = 2r + 2

answer is A.

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Detailed Solution

$\begin{array}{l} ^{2 n} C_{3 r - 1} =^{2 n} C_{r + 1} \Rightarrow 3 r - 1 + r + 1 = 2 n \\ \Rightarrow n = 2 r . \end{array}$

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