Determine the ordinal number of the term of the expansion of $\left(\dfrac{3}{4} \sqrt[3]{a^2} + \dfrac{2}{3} \sqrt{a}\right)^{12}$ which contains $a^7$.
In the expansion of $\left(x+y\right)^n$, $\;$ $\left(r+1\right)^{th}$ term $= T_{r+1} = C^n_r \; x^{n-r} \; y^r$
$\left(r+1\right)^{th}$ term in the expansion of $\left(\dfrac{3}{4} \sqrt[3]{a^2} + \dfrac{2}{3} \sqrt{a}\right)^{12} = \left(\dfrac{3}{4} a^{\frac{2}{3}} + \dfrac{2}{3} a^{\frac{1}{2}}\right)^{12}$ is
$T_{r+1} = C^{12}_r \; \left(\dfrac{3}{4} a^{\frac{2}{3}}\right)^{12-r} \; \left(\dfrac{2}{3} a^{\frac{1}{2}}\right)^r$
i.e. $\;$ $T_{r+1} = C^{12}_r \times \dfrac{3^{12-r} \; a^{\frac{24-2r}{3}}}{2^{24-2r}} \times \dfrac{2^r \; a^{\frac{r}{2}}}{3^r}$
i.e. $\;$ $T_{r+1} = C^{12}_r \times \dfrac{3^{12-2r} \; a^{\frac{24-2r}{3} + \frac{r}{2}}}{2^{24-3r}}$
Let the $\left(r+1\right)^{th}$ term in the expansion contain $a^7$.
Then,
$\dfrac{24-2r}{3} + \dfrac{r}{2} = 7$
i.e. $\;$ $48 - 4r + 3r = 42$
i.e. $\;$ $r = 6$
$\therefore \;$ The ordinal number of the term in the expansion of $\left(\dfrac{3}{4} \sqrt[3]{a^2} + \dfrac{2}{3} \sqrt{a}\right)^{12}$ which contains $a^7$ is $6$.