<s>[INST] What is the coefficient of the term $x^4 y^5$ in $(x+y+2)^{12}$? What is the coefficient of the term $x^4 y^5$ in $(x+y+2)^{12}$?
How can we calculate this expression  ?
I've applied the binomial theorem formula and got $91$ terms but I am not sure if it is right or wrong.
 [/INST] You could use the binomial theorem twice. Let $[x^{k}]$ denote the coefficient of $x^k$ from the polynomial $P(x)=\sum_{j=0}^{n}a_jx^j$, i.e. $[x^{k}]P(x)=a_{k}$.
Now, $$\begin{eqnarray}[x^4y^5](x+y+2)^{12}&=&[x^4y^5](x+(y+2))^{12}\\&=&[x^4y^5]\sum_{j=0}^{12}\binom{12}{j}x^j(y+2)^{12-j}\\&=&[y^5]\binom{12}{4}(y+2)^{12-4}\\&=&\binom{12}{4}[y^5](y+2)^8\\&=&\binom{12}{4}[y^5]\sum_{j=0}^{8}\binom{8}{j}y^j2^{8-j}\\&=&\binom{12}{4}\binom{8}{5}2^{8-5}\\&=&\frac{12!}{4!8!}\frac{8!}{5!3!}2^3=\frac{12!2^3}{3!4!5!}\end{eqnarray}$$
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