<s>[INST] How many terms are there containing the term $xyk^2$ in the expansion of $(2x-y+t+3z+4k)^8$ 
How many terms are there containing the term $xyk^2$ in the expansion of $(2x-y+t+3z+4k)^8$ such as $xyk^2t^2z^2$ or $xyk^2t^4z^0$ or $xyk^2z^3t$ etc.

I made up this question and calculated it , but i do not know whether my solution is correct or not.
$\color{blue}{Solution:}$
$\color{red}{1-)}$ I turned it out to be binomial expression such that $\color{green}{a=}2x-y+4k$ and $\color{green}{b=}t+3z$ , so I will work over $(a+b)^8$.
$\color{red}{2-)}$ I thought that i should look for  $xyk^2$ in $a$  ,so i said that the exponential of $a$ must be $4$ , because $1+1+2=4$  which are the sum of the exponents of  $xyk^2$.
$\color{red}{3-)}$ Hence, the exponent of $b$ must be $4$ ,as well. Because the sum of the exponents of $a$ and $b$ must be equal to $8$.
$\color{red}{4-)}$ Therefore, the terms which contains $xyk^2$ in the expansion of $(2x-y+t+3z+4k)^8$ must be equal the numbers of terms in the expansion of $b$ ,i.e, $(3z+t)^4$. It is equal to $5$
So , my answer is $5$ . I checked it with second solution. However , i am not sure about the truth of the first way. Is my first way true?
Thanks for all contributions..
$\color{blue}{Solution 2:}$ Because of the fact that the sum of the exponents must be equal to $8$ , i should find the number of partition of $4$ in two part so as to find the number of combinations of the exponents of $t$ and $z$. Then, $t=4,z=0$ or $t=3,z=1$ or $t=2,z=2$ or $t=1,z=3$ or $t=0,z=4$
 [/INST] Both solutions are correct. The reasoning to obtain $5$ in the first solution is somewhat more detailed. The essence is in both solutions the same, namely to keep the focus on the number of different terms in the expansion of $(t+3z)^4$.
Here is a crosscheck (kinda overkill). We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance
\begin{align*}
[x^k](ax+z)^n=\binom{n}{k}a^kz^{n-k}\tag{1}
\end{align*}

We obtain by successively applying (1)
\begin{align*}
\color{blue}{[xyk^2]}&\color{blue}{(2x-y+t+3z+4k)^8}\\
&=\binom{8}{1}2^1[yk^2](-y+t+3z+4k)^7\\
&=\binom{8}{1}2^1\binom{7}{1}(-1)^1[k^2](t+3z+4k)^6\\
&=\binom{8}{1}2^1\binom{7}{1}(-1)^1\binom{6}{2}4^2(t+3z)^4\\
&\,\,\color{blue}{=-26\,880(t+3z)^4}\\
\end{align*}
The number of wanted terms is the number of different terms of $(t+3z)^4$ expanded, which is $5$.

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