# What is the coefficient of x^{8}y^{9} in the expansion of (

Liesehf 2021-11-21 Answered
What is the coefficient of
$$\displaystyle{x}^{{{8}}}{y}^{{{9}}}$$
in the expansion of
$$\displaystyle{\left({3}{x}+{2}{y}\right)}^{{{17}}}$$?

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## Expert Answer

Leory2000
Answered 2021-11-22 Author has 673 answers
Setp 1
Binomial theorem
$$\displaystyle{\left({x}+{y}\right)}^{{{n}}}=\sum^{{{n}}}_{\left\lbrace{j}={0}\right\rbrace}$$ nj $$\displaystyle{x}^{{{n}-{j}}}{y}^{{{j}}}$$
We are interested in the term $$\displaystyle{x}^{{{8}}}{y}^{{{9}}}\in{P}{S}{K}{\left({3}{x}+{2}{y}\right)}^{{{17}}}$$
n=17
j=9
The coresponding term is then:
$$\displaystyle{\left({n},{j}\right)}{\left({3}{x}\right)}^{{{n}-{j}}}{\left({2}{y}\right)}^{{{j}}}={\left({17},{9}\right)}{\left({3}{x}\right)}^{{{17}-{9}}}{\left({2}{y}\right)}^{{{9}}}$$
$$\displaystyle={\frac{{{17}!}}{{{9}!{\left({17}-{9}!\right)}}}}{\left({3}{x}\right)}^{{{8}}}{\left({2}{y}\right)}^{{{9}}}$$
$$\displaystyle={\frac{{{17}!}}{{{9}!{8}!}}}{3}^{{{8}}}{x}^{{{8}}}{2}^{{{9}}}{y}^{{{9}}}$$
$$\displaystyle={24310}\cdot{3}^{{{8}}}{2}^{{{9}}}{x}^{{{8}}}{y}^{{{9}}}$$
$$\displaystyle={81},{662},{929},{920}{x}^{{{8}}}{y}^{{{9}}}$$
Thus the coefficient of $$\displaystyle{x}^{{{8}}}{y}^{{{9}}}$$is then 81, 662, 929, 920
So,
81, 662, 929, 920

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