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If the coefficients of ${(r - 5)}^{t h}$ and ${(2 r - 1)}^{t h}$ terms in the expansion of ${(1 + x)}^{34}$ are equal then $r =$

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answer is C.

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Detailed Solution

Coefficient of $T_{r - 5} =$ coefficient of $T_{2 r - 1}$
$\begin{array}{l} ^{34} C_{r - 6} =^{34} C_{2 r - 2} ∴ r - 6 = 2 r - 2 (or) 34 = r - 6 + 2 r - 2 = 4 \\ 3 r = 42 \Rightarrow r = 14 \end{array}$

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