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I need help finding the coefficient of $x^7y^2$ in the expansion of $(2x-y)^9$ if you could give me a hint

user124557
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2 Answers2

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Hint: The coefficient of $s^7t^2$ in the expansion of $(s+t)^9$ is $\binom{9}{7}$. Let $s=2x$ and $t=-y$.

André Nicolas
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  • I understand $\binom{9}{7}$ but I'm confused with $s=2x$ part – user124557 Feb 02 '14 at 21:11
  • Keep in mind that the terms in your binomial are $ \ 2x \ $ and $ \ -y \ $ , so the term having the variable powers $ \ x^7 y^2 \ $ is going to be the consequence of multiplying out the binomial and having a term which has $ \ (2x)^7 \ \cdot \ (-y)^2 \ . $ This will then be multiplied by the combinatorial factor $ \ \binom{9}{7} \ .$ – colormegone Feb 02 '14 at 22:14
  • We will replace $s$ by $2x$ and $t$ by $-y$. The $(2x)^7$ is $2^7 x^7$. The $(-y)^2$ is plain $y^2$. So the coefficient of $x^7y^2 is $2^7\binom{9}{7}$. We may wish to "simplify." – André Nicolas Feb 02 '14 at 23:15
  • Thanks, I was able to get it. $2^7 * \binom{9}{7} = 4608$ – user124557 Feb 02 '14 at 23:26
  • You are welcome. – André Nicolas Feb 03 '14 at 00:16
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It's the appropriate binomial coefficient times two to the seventh... and what's the sign?

Pifagor
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