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 ?
 A: 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*}
A: Here's a hint to simplify the calculation. Suppose you were to fully expand $(x^2+y+2t+3k)^{10}$ and then drop all but those terms containing $x^2 t$. What results is an expression of the form
$$ax^2 t+bk x^2 t+cy x^2 t+dk^2 x^2 t+\cdots=(a+bk+cy+dk^2+\cdots)x^2 t$$
Summing up these coefficients yields $a+b+c +d+\cdots$.
What's the relation between this sum and the polynomial in $y,k$ written above? (This reduces the problem to finding the coefficient of $x^2 t$ in some polynomial in $x,t$.)
A: As an alternative, we can use the multinomial expansion formula:
$$(x_1+x_2+\cdots +x_m)^n=\sum_{i_1+i_2+\cdots +i_m=n} {n\choose i_1,i_2,...,i_m} \prod_{t=1}^m x_t^{i_t}, \\
\quad \text{where} \quad {n\choose i_1,i_2,...,i_m}=\frac{n!}{i_1!i_2!\cdots i_m!}\\
 $$
$$ (x^2 +y +2t +3k)^{10}=\sum_{i_1+i_2+i_3+i_4=10} {10\choose i_1,i_2,i_3,i_4} x^{2i_1}y^{i_2}(2t)^{i_3}(3k)^{i_4}$$
We want to have $i_1=1,i_3=3$:
$$2^33^6{10\choose 1,0,3,6}+2^33^5{10\choose 1,1,3,5}+2^33^4{10\choose 1,2,3,4}+\\
2^33^3{10\choose 1,3,3,3}+2^33^2{10\choose 1,4,3,2}+2^33^1{10\choose 1,5,3,1}+2^3{10\choose 1,6,3,0}= \\
2^3\cdot \left[\frac{10!}{3!6!}(3^6+1)+\frac{10!}{3!5!}(3^5+3)+\frac{10!}{2!3!4!}(3^4+3^2)+3^3\frac{10!}{3!3!3!}\right]=\\
2^3\cdot \left[613,200+1,239,840+1,134,000+453,600\right]=\\
27,525,120.$$
