check my work

**papa_chango123** · December 9th 2006, 07:38 AM

I am to evaluate the following:

C(7,2)
answer:21

C(12,7)
answer: 66

This is the one I can't figure out

the coefficient of x7y2 in the expansion of (2x-y)9

any help is appreciated

**Soroban** · December 9th 2006, 08:20 AM

Hello, papa_chango123!

Evaluate: . $C(7,2)$
. . Answer: 21 . . . Right!

$C(12,7)$
. . Answer: 66 . . . no

$C(12,7) \:=\:\frac{12!}{7!5!} \:=\:\frac{12\cdot11\cdot10\cdot9\cdot8}{5\cdot4\c dot3\cdot2\cdot1} \:=\:792$

Find the coefficient of $x^7y^2$ in the expansion of $(2x-y)^9$

You're expected to know the Binomial Expansion.

The term with $x^7y^2$ is: . $C(9,7)(2x)^7(-y)^2$

Hence, we have: . $\frac{9!}{7!2!}(2^7x^7)(y^2)\;=\;\boxed{4608}x^7y^ 2$

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