Question:

Step 1: Find the term independent of $x$ in the expansion of $\left(3x - \dfrac{1}{x}\right)^{20}$. The general term in the binomial expansion is: $$T_{r+1} = \binom{20}{r}(3x)^{20-r}\left(-\dfrac{1}{x}\right)^r = \binom{20}{r}3^{20-r}(-1)^r x^{20-2r}$$ For the term to be independent of $x$, the power of $x$ must be zero: $$20 - 2r = 0 \implies r = 10$$ Therefore: $$A = \binom{20}{10}3^{10}(-1)^{10} = \binom{20}{10}3^{10}$$ Step 2: Find the term independent of $x$ in the expansion of $\left(x + \dfrac{\sqrt[9]{3^{10}}}{x}\right)^{18}$. First, simplify the second term: $\sqrt[9]{3^{10}} = 3^{10/9}$ The general term in the binomial expansion is: $$T_{r+1} = \binom{18}{r}x^{18-r}\left(\dfrac{3^{10/9}}{x}\right)^r = \binom{18}{r}3^{10r/9}x^{18-2r}$$ For the term to be independent of $x$: $$18 - 2r = 0 \implies r = 9$$ Therefore: $$B = \binom{18}{9}3^{10 \cdot 9/9} = \binom{18}{9}3^{10}$$ Step 3: Simplify $\dfrac{9}{38}A + B$. $$\dfrac{9}{38}A + B = \dfrac{9}{38}\binom{20}{10}3^{10} + \binom{18}{9}3^{10} = 3^{10}\left[\dfrac{9}{38}\binom{20}{10} + \binom{18}{9}\right]$$ Step 4: Simplify $\dfrac{9}{38}\binom{20}{10}$. Express $\binom{20}{10}$ in terms of $\binom{18}{8}$: $$\binom{20}{10} = \dfrac{20!}{10! \cdot 10!} = \dfrac{20 \cdot 19 \cdot 18!}{10 \cdot 9 \cdot 8! \cdot 10!} = \dfrac{20 \cdot 19}{90}\binom{18}{8}$$ Therefore: $$\dfrac{9}{38}\binom{20}{10} = \dfrac{9}{38} \cdot \dfrac{20 \cdot 19}{90}\binom{18}{8} = \dfrac{9 \cdot 380}{38 \cdot 90}\binom{18}{8} = \dfrac{3420}{3420}\binom{18}{8} = \binom{18}{8}$$ Step 5: Use Pascal's identity to combine the binomial coefficients. $$\dfrac{9}{38}A + B = 3^{10}\left[\binom{18}{8} + \binom{18}{9}\right]$$ By Pascal's identity: $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$ $$\binom{18}{8} + \binom{18}{9} = \binom{19}{9}$$ Therefore: $$\dfrac{9}{38}A + B = 3^{10}\binom{19}{9} = 3^{10} \cdot {}^{19}C_9$$ The answer is **Option 2: $3^{10} \cdot {}^{19}C_9$**