# Binomial Cooeficient help check

#### kory

##### Junior Member
Without expanding the entire expression, find the coefficient on the $$\displaystyle x^{12}y^5$$ term for the binomial expansion of $$\displaystyle (2x-3y)^{17}$$

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?
$$\displaystyle _{17}C_5(2x)^{12}(-3y)^5$$

$$\displaystyle 17! / 12! 5!$$

$$\displaystyle 4096x^{12}(-243)y^5$$

$$\displaystyle 17 * 16 * 15* 14* 13$$ / $$\displaystyle 5 * 4* 3* 2$$

$$\displaystyle 6188 * 4096 * -243 = -6,159,089,664x^{12}y^5$$

Is this still considered expanding?

#### lex

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

#### JeffM

##### Elite Member
I’d probably just write

$$\displaystyle \dbinom{17}{5} (2x)^{12}(-3y)^5 = - \dfrac{17 * 16 * 15 * 14 * 13}{5 * 4 * 3 * 2} * 2^{12} * 3^{5} x^{12}y^5 = - 17 * 13 * 7 * 3^5 * 2^{14} x^{12} y^{5}.$$

Pay attention to parentheses. With that kind of number, we might find prime factorization useful.

#### kory

##### Junior Member
Good to know. Thank you