<s>[INST] What is the coefficient of $x^3 y^4$ in the expansion of $ (2x-y+5)^8$ I was thinking of doing $\binom{8}{4}$ but not sure if right. 
 [/INST] Pre-expansion, there are $8$ factors of $2x - y + 5$.
From those $8$ factors, choose the $3$ that contribute to the $x^3$, from the remaining $5$ factors, choose the $4$ that will contribute to $y^4$.  There is only 1 factor left so it chooses itself.
Post-expansion, that's ${8 \choose 3}{5 \choose 4}{1 \choose 1}$ terms with $x^3y^4$, and taking into account the initial coefficients of $\{2, -1, 5\}$, the final coefficient is:
$${8 \choose 3}{5 \choose 4}2^3(-1)^45^1 = 11200$$
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