<s>[INST] What is the sum of the coefficients of the terms containing $x^2t^3$ in the given expansion. 
It is given that $(x^2 +y +2t +3k)^{10}$. What is the sum of the coefficients of the terms containing $x^2t^3$  in the given expansion. For example the terms are $x^2y^4t^3k^2$ etc.

I said that let $(x^2 +2t)$ be $"a"$ and $(y +3k)$ be $"b"$. Then, $(x^2 +y +2t +3k)^{10}=(a+b)^{10}$.
If we are looking for  $x^2t^3$, then the exponential of $a$ must be $4$. Then, the exponential of $b$ is $6$.
Then, the coefficient of $x^2t^3$ is $4 \times 2^3=32$. Now, we have $(y +3k)^6$ as $b$. If we find the sum of the all coefficients in $(y +3k)^6$, we can handle the question.
The sum of the all coefficients in $(y +3k)^6$  is $4^6$.
Then, answer is $32 \times 4^6$
Is my solution correct ?
 [/INST] Following OPs approach somewhat more formally we obtain
\begin{align*}
\color{blue}{[x^2t^3]}&\color{blue}{\left(\left(x^2+2t\right)+(y+3k)\right)^{10}}\\
&=[x^2t^3]\sum_{q=0}^{10}\binom{10}{q}\left(x^2+2t\right)^q\left(y+3k\right)^{10-q}\\
&=\sum_{q=0}^{10}\binom{10}{q}\left([x^2t^3]\sum_{r=0}^q\binom{q}{r}x^{2r}(2t)^{q-r}\right)(y+3k)^{10-q}\tag{1}\\
&=\binom{10}{4}\binom{4}{1}2^3(y+3k)^6\\
&\,\,\color{blue}{=6\,720\cdot(y+3k)^6}\tag{2}
\end{align*}
We see in (1) in order to select the coefficient of $x^2$ and $t^3$ we have to choose $r=1$ and $q=4$. Evaluating (2) at $y=1$ and $k=1$ we obtain
\begin{align*}
\color{blue}{6\,720\cdot4^6}
\end{align*}
</s>