Binomial Cooeficient help check

[kory]

Without expanding the entire expression, find the coefficient on the [math]x^{12}y^5[/math] term for the binomial expansion of [math](2x-3y)^{17}[/math]
Normally I would create a tree and then simplify a long expression for these types of problems but this one says to do it without expanding.

So I'm assuming I would I do something like this?
[math]_{17}C_5(2x)^{12}(-3y)^5[/math]
[math]17! / 12! 5![/math]
[math]4096x^{12}(-243)y^5[/math]
[math]17 * 16 * 15* 14* 13[/math] / [math]5 * 4* 3* 2[/math]
[math]6188 * 4096 * -243 = -6,159,089,664x^{12}y^5[/math]
Is this still considered expanding?

 

[lex]

That's perfect. The coefficient is just the number though.
(It only said without expanding the entire expression).

 

[JeffM]

I’d probably just write

[MATH]\dbinom{17}{5} (2x)^{12}(-3y)^5 = - \dfrac{17 * 16 * 15 * 14 * 13}{5 * 4 * 3 * 2} * 2^{12} * 3^{5} x^{12}y^5 = - 17 * 13 * 7 * 3^5 * 2^{14} x^{12} y^{5}.[/MATH]
Pay attention to parentheses. With that kind of number, we might find prime factorization useful.

 

[kory]

Good to know. Thank you